CBSE Class 12 Mathematics Continuity And Differentiability Assignment Set 01

CBSE Class 12 Mathematics Continuity And Differentiability Assignment Set A. Chapter wise assignments are being given by teachers to students to make them understand the chapter concepts. Its extremely critical for all CBSE students to practice all assignments which will help them in gaining better marks in examinations. All assignments available for free download on the website are developed by the best teachers having many years of teaching experience in CBSE schools all over the country. Students, teachers and parents are advised to download the CBSE study material.

Points to Remember

• A function \( f(x) \) is said to be continuous at \( x = c \) if and only if \( \lim_{x \to c} f(x) = f(c) \).
  This means: \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \).
• \( f(x) \) is continuous in the open interval \( (a, b) \) if and only if it is continuous at every \( x = c \) for all \( c \in (a, b) \).
• \( f(x) \) is continuous in the closed interval \( [a, b] \) if and only if:
  (i) \( f(x) \) is continuous in \( (a, b) \)
  (ii) \( \lim_{x \to a^+} f(x) = f(a) \)
  (iii) \( \lim_{x \to b^-} f(x) = f(b) \).
• All trigonometric functions are continuous on their respective domains.
• Every polynomial function is continuous on the set of real numbers \( \mathbb{R} \).
• If \( f(x) \) and \( g(x) \) are two continuous functions and \( c \in \mathbb{R} \), then at \( x = a \):
  (i) \( f(x) \pm g(x) \) are also continuous at \( x = a \).
  (ii) \( g(x) \cdot f(x) \), \( f(x) + c \), \( cf(x) \), and \( |f(x)| \) are also continuous at \( x = a \).
  (iii) \( \frac{f(x)}{g(x)} \) is continuous at \( x = a \), provided that \( g(a) \ne 0 \).
• \( f(x) \) is derivable (differentiable) at \( x = c \) in its domain if and only if:
  \[ \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \] and this limit is finite. This value is denoted by \( f'(c) \).

Standard Derivatives

• Product Rule: \( \frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx} \)
• Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
• Chain Rule: If \( y = f(u) \) and \( u = g(t) \), then \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = f'(u) \cdot g'(t) \)
• Inverse Trigonometric Derivatives:
  \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \), \( \frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}} \)
  \( \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2} \), \( \frac{d}{dx}(\cot^{-1} x) = \frac{-1}{1+x^2} \)
  \( \frac{d}{dx}(\sec^{-1} x) = \frac{1}{x\sqrt{x^2-1}} \), \( \frac{d}{dx}(\csc^{-1} x) = \frac{-1}{x\sqrt{x^2-1}} \)
• Exponential and Logarithmic Derivatives:
  \( \frac{d}{dx}(e^x) = e^x \), \( \frac{d}{dx}(\log x) = \frac{1}{x} \)
• The greatest integer function \( f(x) = [x] \) is discontinuous at all integral points, but continuous for all non-integer real values \( x \in \mathbb{R} - \mathbb{Z} \).
• Rolle's Theorem: If \( f(x) \) is continuous in \( [a, b] \), differentiable in \( (a, b) \), and \( f(a) = f(b) \), then there exists at least one real number \( c \in (a, b) \) such that \( f'(c) = 0 \).
• Mean Value Theorem: If \( f(x) \) is continuous in \( [a, b] \) and differentiable in \( (a, b) \), then there exists at least one real number \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
• \( f(x) = \log_e x \) is a continuous function for all \( x > 0 \).

Very Short Answer Type Questions (1 Mark)

Question 1. For what value of x, f(x) = |2x - 7| is not derivable.
Answer:
A modular function of the form \( f(x) = |g(x)| \) is not differentiable (derivable) at the points where \( g(x) = 0 \).
Setting the expression inside the modulus to zero:
\( 2x - 7 = 0 \implies x = \frac{7}{2} \).
Therefore, \( f(x) \) is not derivable at \( x = \frac{7}{2} \).
In simple words: The absolute value graph has a sharp corner (v-shape) where the inner term equals zero. Since a curve is not differentiable at a sharp corner, the function is not derivable at \( x = 7/2 \).

Exam Tip: For any function \( |ax + b| \), the point of non-derivability is always the root of the linear expression, which is \( x = -b/a \).

Question 2. Write the set of points of continuity of g(x) = |x - 1| + |x + 1|.
Answer:
The function \( g(x) \) is the sum of two modulus functions, \( |x - 1| \) and \( |x + 1| \).
Since absolute value functions are continuous everywhere on the real number line, their sum must also be continuous on the entire domain of real numbers.
Thus, the set of points of continuity of \( g(x) \) is \( \mathbb{R} \) (the set of all real numbers).
In simple words: Modulus functions are continuous everywhere. Adding two continuous functions together always results in another continuous function, so it is continuous for all real numbers.

Exam Tip: Do not confuse continuity with differentiability; absolute value functions are continuous everywhere, even though they are not differentiable at their corner points.

Question 3. What is derivative of |x - 3| at x = - 1.
Answer:
For \( x < 3 \), the expression \( |x - 3| \) is negative, so we rewrite it as:
\( |x - 3| = -(x - 3) = -x + 3 \).
Since we are evaluating the derivative at \( x = -1 \), which is strictly less than 3, we differentiate the simplified expression:
\( \frac{d}{dx}(|x - 3|) = \frac{d}{dx}(-x + 3) = -1 \).
Therefore, the derivative at \( x = -1 \) is \( -1 \).
In simple words: Since \( x = -1 \) is less than 3, the absolute value is negative, making the function \( -x + 3 \). The derivative of this straight line is simply its slope, which is \( -1 \).

Exam Tip: Always redefine modular functions for the specific neighborhood of the point where you want to calculate the derivative before differentiating.

Question 4. What are the points of discontinuity of f(x) = \( \frac{(x - 1) + (x + 1)}{(x - 7)(x - 6)} \).
Answer:
A rational function of the form \( f(x) = \frac{p(x)}{q(x)} \) is discontinuous at the points where its denominator is equal to zero.
Setting the denominator to zero:
\( (x - 7)(x - 6) = 0 \implies x = 7 \) or \( x = 6 \).
Therefore, the points of discontinuity of \( f(x) \) are \( x = 6 \) and \( x = 7 \).
In simple words: A fraction is not defined when its bottom part becomes zero. Since the function is not defined at those points, it is discontinuous there.

Exam Tip: Rational functions are continuous everywhere on their domains. Their points of discontinuity are simply the real roots of the denominator polynomial.

Question 5. Write the number of points of discontinuity of f(x) = [x] in [3, 7].
Answer:
The greatest integer function \( [x] \) is discontinuous at all integer points because the value jumps abruptly at every integer.
In the closed interval \( [3, 7] \), the integer values are \( 3, 4, 5, 6, 7 \).
- At the left boundary \( x = 3 \): the function is continuous from the right because \( \lim_{x \to 3^+} [x] = 3 = f(3) \).
- At the right boundary \( x = 7 \): the function is discontinuous because the left-hand limit \( \lim_{x \to 7^-} [x] = 6 \), which does not equal \( f(7) = 7 \).
- At the interior integer points \( x = 4, 5, 6 \): the function is discontinuous because the left-hand and right-hand limits are unequal.
Thus, the points of discontinuity in the interval \( [3, 7] \) are \( x = 4, 5, 6, 7 \).
The total number of points of discontinuity is 4.
In simple words: The step function jumps at every integer. In the interval from 3 to 7, these jumps happen at 4, 5, 6, and 7, making a total of 4 discontinuous points.

Exam Tip: Pay close attention to closed interval endpoints. The greatest integer function is continuous from the right at the left endpoint, but discontinuous at the right endpoint.

Question 6. The function, f(x) = \( \begin{cases} \lambda x - 3 & \text{if } x < 2 \\ 4 & \text{if } x = 2 \\ 2x & \text{if } x > 2 \end{cases} \) is a continuous function for all x \in R, find \( \lambda \).
Answer:
Since the function \( f(x) \) is continuous everywhere, it must be continuous at the boundary point \( x = 2 \).
This requires that:
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \).
Calculate the left-hand limit (LHL):
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2} (\lambda x - 3) = 2\lambda - 3 \).
Calculate the right-hand limit (RHL):
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2} (2x) = 4 \).
Given \( f(2) = 4 \), we set:
\( 2\lambda - 3 = 4 \)
\( 2\lambda = 7 \implies \lambda = \frac{7}{2} \).
In simple words: For the graph to connect smoothly without any break at \( x = 2 \), the left-side equation value must match the right-side limit value of 4, which gives \( \lambda = 7/2 \).

Exam Tip: Set up the LHL = RHL = value condition clearly, as examiners award individual step marks for stating this continuity criterion.

Question 7. For what value of K, f(x) = \( \begin{cases} \frac{\tan 3x}{\sin 2x}, & x \ne 0 \\ 2K, & x = 0 \end{cases} \) is continuous \( \forall x \in R \).
Answer:
For \( f(x) \) to be continuous at \( x = 0 \), we must have:
\( \lim_{x \to 0} f(x) = f(0) \).
Let us evaluate the limit:
\( \lim_{x \to 0} \frac{\tan 3x}{\sin 2x} = \lim_{x \to 0} \left[ \frac{\left(\frac{\tan 3x}{3x}\right) \cdot 3x}{\left(\frac{\sin 2x}{2x}\right) \cdot 2x} \right] \)
\( = \frac{3}{2} \cdot \frac{\lim_{x \to 0} \frac{\tan 3x}{3x}}{\lim_{x \to 0} \frac{\sin 2x}{2x}} \).
Since \( \lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1 \) and \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \):
\( = \frac{3}{2} \cdot \frac{1}{1} = \frac{3}{2} \).
Now set this limit equal to the value of the function at \( x = 0 \):
\( 2K = \frac{3}{2} \implies K = \frac{3}{4} \).
In simple words: Use standard limits to find that the fractional part approaches \( 3/2 \) as \( x \) goes to 0. Equating this limit to \( 2K \) gives \( K = 3/4 \).

Exam Tip: Show the division and multiplication of angles (\( 3x \) and \( 2x \)) explicitly to justify your limit transition steps clearly.

Question 8. Write derivative of sin x w.r.t. cos x.
Answer:
Let \( u = \sin x \) and \( v = \cos x \).
We want to find the derivative of \( u \) with respect to \( v \), which is \( \frac{du}{dv} \).
Using parametric differentiation:
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} \).
Since \( \frac{du}{dx} = \cos x \) and \( \frac{dv}{dx} = -\sin x \):
\( \frac{du}{dv} = \frac{\cos x}{-\sin x} = -\cot x \).
In simple words: Differentiate both functions with respect to \( x \) first, then divide the derivative of the first function by the derivative of the second function to get \( -\cot x \).

Exam Tip: Always use the formula \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \) for derivative of one function with respect to another function.

Question 9. If f(x) = \( x^2 g(x) \) and g(1) = 6, g´(1) = 3 find value of f´(1).
Answer:
Applying the product rule of differentiation to \( f(x) \):
\( f'(x) = \frac{d}{dx}(x^2) \cdot g(x) + x^2 \cdot \frac{d}{dx}(g(x)) \)
\( f'(x) = 2x \cdot g(x) + x^2 \cdot g'(x) \).
Now substitute \( x = 1 \):
\( f'(1) = 2(1) \cdot g(1) + (1)^2 \cdot g'(1) \)
\( f'(1) = 2 \cdot (6) + 1 \cdot (3) = 12 + 3 = 15 \).
In simple words: Use the product rule to write out the general derivative equation, and then plug in the given values at \( x = 1 \) to compute the final answer.

Exam Tip: Be sure to write the general derivative expression with \( x \) first before substituting \( x = 1 \) to prevent operational calculation errors.

Question 10. Write the derivative of the following functions :
(i) \( \log_3(3x + 5) \)
(ii) \( e^{\log_2 x} \)
(iii) \( e^{6 \log_e(x-1)} \), \( x > 1 \)
(iv) \( \sec^{-1}\sqrt{x} + \csc^{-1}\sqrt{x} \), \( x \ge 1 \).
(v) \( \sin^{-1}(x^{7/2}) \)
(vi) \( \log_x 5 \), \( x > 0 \).
Answer:
(i) Using the base change formula, rewrite the function as \( \frac{\ln(3x + 5)}{\ln 3} \).
Differentiating using chain rule:
\( \frac{d}{dx} \left[ \frac{\ln(3x + 5)}{\ln 3} \right] = \frac{1}{\ln 3} \cdot \frac{1}{3x + 5} \cdot 3 = \frac{3}{(3x + 5)\ln 3} \).

(ii) We rewrite the exponent term using natural logarithms: \( \log_2 x = \frac{\ln x}{\ln 2} \).
So, \( e^{\log_2 x} = e^{\ln x / \ln 2} \).
Differentiating using the chain rule:
\( \frac{d}{dx} \left( e^{\ln x / \ln 2} \right) = e^{\ln x / \ln 2} \cdot \frac{d}{dx} \left( \frac{\ln x}{\ln 2} \right) = e^{\log_2 x} \cdot \frac{1}{x \ln 2} \).

(iii) Using logarithmic simplification:
\( e^{6\log_e(x-1)} = e^{\log_e(x-1)^6} = (x - 1)^6 \).
Differentiating this simple power function:
\( \frac{d}{dx}\left[ (x - 1)^6 \right] = 6(x - 1)^5 \).

(iv) Since \( \sec^{-1} u + \csc^{-1} u = \frac{\pi}{2} \) for all valid inputs, we can simplify:
\( \sec^{-1}\sqrt{x} + \csc^{-1}\sqrt{x} = \frac{\pi}{2} \).
The derivative of a constant is zero, so:
\( \frac{d}{dx}\left( \frac{\pi}{2} \right) = 0 \).

(v) Differentiating using the chain rule:
\( \frac{d}{dx} \left[ \sin^{-1}(x^{7/2}) \right] = \frac{1}{\sqrt{1 - (x^{7/2})^2}} \cdot \frac{d}{dx}(x^{7/2}) \)
\( = \frac{1}{\sqrt{1 - x^7}} \cdot \frac{7}{2}x^{5/2} = \frac{7x^{5/2}}{2\sqrt{1 - x^7}} \).

(vi) Rewrite the function using change-of-base rule: \( \log_x 5 = \frac{\ln 5}{\ln x} = (\ln 5)(\ln x)^{-1} \).
Differentiating:
\( \frac{d}{dx}\left[ (\ln 5)(\ln x)^{-1} \right] = (\ln 5) \cdot (-1)(\ln x)^{-2} \cdot \frac{1}{x} = -\frac{\ln 5}{x(\ln x)^2} \).
In simple words: Simplify exponential and logarithmic functions using standard properties before differentiating. For inverse trigonometric sums, note they sum to constants whose derivatives are 0.

Exam Tip: Simplifying functions algebraically (like rewriting \( e^{6\log(x-1)} \) as \( (x-1)^6 \)) saves you from doing long and messy chain-rule derivatives.

Short Answer Type Questions (4 Marks)

Question 11. Discuss the continuity of following functions at the indicated points.
(i) \( f(x) = \begin{cases} \frac{x - |x|}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases} \) at \( x = 0 \).
(ii) \( g(x) = \begin{cases} \frac{\sin 2x}{3x}, & x \ne 0 \\ \frac{3}{2}, & x = 0 \end{cases} \) at \( x = 0 \).
(iii) \( f(x) = \begin{cases} x^2 \cos(1/x), & x \ne 0 \\ 0, & x = 0 \end{cases} \) at \( x = 0 \).
(iv) \( f(x) = |x| + |x - 1| \) at \( x = 1 \).
(v) \( f(x) = \begin{cases} x - [x], & x \ne 1 \\ 0, & x = 1 \end{cases} \) at \( x = 1 \).
Answer:
(i) Let us find the left-hand limit (LHL) and right-hand limit (RHL) at \( x = 0 \):
- For LHL (\( x \to 0^- \)): Since \( x < 0 \), we have \( |x| = -x \).
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x - (-x)}{x} = \lim_{x \to 0^-} \frac{2x}{x} = 2 \).
- For RHL (\( x \to 0^+ \)): Since \( x > 0 \), we have \( |x| = x \).
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x - x}{x} = \lim_{x \to 0^+} 0 = 0 \).
Since LHL (\( 2 \)) \( \ne \) RHL (\( 0 \)), the limit of the function as \( x \to 0 \) does not exist.
Therefore, \( f(x) \) is discontinuous at \( x = 0 \).

(ii) Let us find the limit of \( g(x) \) as \( x \to 0 \):
\( \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{\sin 2x}{3x} = \lim_{x \to 0} \left[ \frac{\sin 2x}{2x} \cdot \frac{2x}{3x} \right] = \frac{2}{3} \lim_{x \to 0} \frac{\sin 2x}{2x} = \frac{2}{3} \cdot 1 = \frac{2}{3} \).
However, the given functional value at \( x = 0 \) is \( g(0) = \frac{3}{2} \).
Since \( \lim_{x \to 0} g(x) \ne g(0) \), the function \( g(x) \) is discontinuous at \( x = 0 \).

(iii) To evaluate \( \lim_{x \to 0} x^2 \cos(1/x) \), we use the Squeeze Theorem.
Since \( -1 \le \cos(1/x) \le 1 \) for all \( x \ne 0 \), we can multiply through by the positive term \( x^2 \):
\( -x^2 \le x^2 \cos(1/x) \le x^2 \).
Evaluating limits on the bounds:
\( \lim_{x \to 0} (-x^2) = 0 \) and \( \lim_{x \to 0} (x^2) = 0 \).
Thus, by Squeeze Theorem, \( \lim_{x \to 0} x^2 \cos(1/x) = 0 \).
Since \( \lim_{x \to 0} f(x) = 0 = f(0) \), the function is continuous at \( x = 0 \).

(iv) Find the LHL, RHL, and functional value at \( x = 1 \) for \( f(x) = |x| + |x - 1| \):
- LHL: \( \lim_{x \to 1^-} (|x| + |x - 1|) = |1| + |0| = 1 \).
- RHL: \( \lim_{x \to 1^+} (|x| + |x - 1|) = |1| + |0| = 1 \).
- Value: \( f(1) = |1| + |1 - 1| = 1 \).
Since LHL = RHL = \( f(1) \), the function is continuous at \( x = 1 \).

(v) Let us find LHL and RHL at \( x = 1 \) for the piecewise function:
- For LHL (\( x \to 1^- \)): Since \( 0 < x < 1 \), we have the greatest integer term \( [x] = 0 \).
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x - [x]) = \lim_{x \to 1^-} (x - 0) = 1 \).
- For RHL (\( x \to 1^+ \)): Since \( 1 < x < 2 \), we have the greatest integer term \( [x] = 1 \).
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - [x]) = \lim_{x \to 1^+} (x - 1) = 1 - 1 = 0 \).
Since LHL (\( 1 \)) \( \ne \) RHL (\( 0 \)), the limit of \( f(x) \) as \( x \to 1 \) does not exist.
Therefore, \( f(x) \) is discontinuous at \( x = 1 \).
In simple words: To check continuity, calculate left and right limits. If the limits are unequal, or if their common limit doesn't match the function's designated value at that point, the graph has a break and is discontinuous.

Exam Tip: For step functions like \( [x] \) or modulus terms like \( |x| \), always write down the specific simplified equations for the left and right neighborhoods before evaluating the limits.

Question 12. For what value of k, f(x) = \( \begin{cases} 3x^2 - kx + 5, & 0 \le x < 2 \\ 1 - 3x, & 2 \le x \le 3 \end{cases} \) is continuous \( \forall x \in [0, 3] \).
Answer:
Since \( f(x) \) is continuous in the closed interval \( [0, 3] \), it must be continuous at the transition boundary \( x = 2 \).
This requires that LHL = RHL = \( f(2) \).
Calculate LHL at \( x = 2 \):
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2} (3x^2 - kx + 5) = 3(2)^2 - k(2) + 5 = 12 - 2k + 5 = 17 - 2k \).
Calculate RHL at \( x = 2 \):
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2} (1 - 3x) = 1 - 3(2) = -5 \).
The functional value is also \( f(2) = 1 - 3(2) = -5 \).
Setting LHL = RHL:
\( 17 - 2k = -5 \)
\( -2k = -5 - 17 \)
\( \implies -2k = -22 \)
\( \implies k = 11 \).
In simple words: The graph must connect smoothly at the transition point \( x = 2 \). Setting the left-hand polynomial equal to the right-hand linear expression at \( x = 2 \) lets us solve for the variable \( k \).

Exam Tip: Be sure to write the limit notation clearly for both LHL and RHL; writing the steps systematically is essential for getting maximum board marks.

Question 13. For what values of a and b \( f(x) = \begin{cases} \frac{x + 2}{|x + 2|} + a & \text{if } x < -2 \\ a + b & \text{if } x = -2 \\ \frac{x + 2}{|x + 2|} + 2b & \text{if } x > -2 \end{cases} \) is continuous at x = -2.
Answer:
For \( f(x) \) to be continuous at \( x = -2 \), we must have:
\( \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = f(-2) \).
First, evaluate the LHL at \( x = -2 \):
Since \( x < -2 \), we have \( x + 2 < 0 \), which implies \( |x + 2| = -(x + 2) \).
\( \lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} \left[ \frac{x + 2}{-(x + 2)} + a \right] = \lim_{x \to -2^-} [-1 + a] = a - 1 \xb \).
Next, evaluate the RHL at \( x = -2 \):
Since \( x > -2 \), we have \( x + 2 > 0 \), which implies \( |x + 2| = x + 2 \).
\( \lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} \left[ \frac{x + 2}{x + 2} + 2b \right] = \lim_{x \to -2^+} [1 + 2b] = 2b + 1 \).
The functional value is given as \( f(-2) = a + b \).
Setting the continuity conditions:
1) LHL = Value:
\( a - 1 = a + b \implies b = -1 \).
2) RHL = Value:
\( 2b + 1 = a + b \).
Substitute \( b = -1 \) into this equation:
\( 2(-1) + 1 = a + (-1) \)
\( -2 + 1 = a - 1 \)
\( -1 = a - 1 \implies a = 0 \).
Therefore, we get \( a = 0 \) and \( b = -1 \).
In simple words: Simplify the fraction in each piece by removing the modulus signs based on the sign of \( x+2 \). Set the resulting left, right, and point values equal to build a system of equations, and solve for \( a \) and \( b \).

Exam Tip: The denominator modulus \( |x+2| \) simplifies to \( -(x+2) \) for \( x < -2 \) and \( +(x+2) \) for \( x > -2 \). Showing this step is load-bearing for proving the limit values.

Short Answer Type Questions (4 Marks)

Question 14. Prove that f(x) = |x + 1| is continuous at x = –1, but not derivable at x = –1.
Answer:
First, let us test the continuity of \( f(x) = |x + 1| \) at \( x = -1 \):
Calculate the left-hand limit (LHL):
\( \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} |x + 1| = |-1 + 1| = 0 \).
Calculate the right-hand limit (RHL):
\( \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} |x + 1| = |-1 + 1| = 0 \).
The functional value at \( x = -1 \) is \( f(-1) = |-1 + 1| = 0 \).
Since LHL = RHL = \( f(-1) \), the function \( f(x) \) is continuous at \( x = -1 \).

Next, let us test the differentiability (derivability) at \( x = -1 \):
Calculate the Left Hand Derivative (LHD) at \( x = -1 \):
\( \text{LHD} = \lim_{h \to 0^-} \frac{f(-1 + h) - f(-1)}{h} = \lim_{h \to 0^-} \frac{|-1 + h + 1| - 0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} \).
Since \( h \to 0^- \), we have \( h < 0 \), which implies \( |h| = -h \):
\( \text{LHD} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \xb \).
Calculate the Right Hand Derivative (RHD) at \( x = -1 \):
\( \text{RHD} = \lim_{h \to 0^+} \frac{f(-1 + h) - f(-1)}{h} = \lim_{h \to 0^+} \frac{|-1 + h + 1| - 0}{h} = \lim_{h \to 0^+} \frac{|h|}{h} \).
Since \( h \to 0^+ \), we have \( h > 0 \), which implies \( |h| = h \):
\( \text{RHD} = \lim_{h \to 0^+} \frac{h}{h} = 1 \).
Since LHD (\( -1 \)) \( \ne \) RHD (\( 1 \)), the function is not derivable at \( x = -1 \).
Thus, the statement is verified.
In simple words: The graph of \( |x+1| \) is continuous because it has no breaks, but it is not derivable at \( x = -1 \) because the graph forms a sharp corner there, making the left and right slopes different.

Exam Tip: Whenever proving non-derivability of modulus functions, clearly show the left-hand and right-hand derivative limits with step-by-step substitution of \( h \).

Question 15. For what value of p, \( f(x) = \begin{cases} x^p \sin(1/x), & x \ne 0 \\ 0, & x = 0 \end{cases} \) is derivable at x = 0.
Answer:
For \( f(x) \) to be derivable at \( x = 0 \), the derivative limit must exist and be finite:
\( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^p \sin(1/h) - 0}{h} = \lim_{h \to 0} h^{p-1} \sin\left(\frac{1}{h}\right) \).
Since \( \sin(1/h) \) oscillates boundedly between \( -1 \) and \( 1 \), the limit \( \lim_{h \to 0} h^{p-1} \sin(1/h) \) will only converge to \( 0 \) if the exponent of \( h \) is strictly positive:
\( p - 1 > 0 \implies p > 1 \).
Therefore, the function is derivable at \( x = 0 \) for all values \( p > 1 \).
In simple words: For the derivative limit to exist at \( x = 0 \), the power of the remaining \( h \) term must be positive so that it squeezes the oscillating sine term to zero as \( h \) approaches 0.

Exam Tip: Remember to state that \( \sin(1/h) \) is a bounded oscillating quantity; this is a necessary justification step in squeeze theorem limits.

Question 16. If \( y = \frac{1}{2} \left[ \tan^{-1}\left( \frac{2x}{1 - x^2} \right) + 2\tan^{-1}\left( \frac{1}{x} \right) \right] \), 0 < x < 1, find \( \frac{dy}{dx} \).
Answer:
Using standard inverse trigonometric relations for \( 0 < x < 1 \):
1) \( \tan^{-1}\left( \frac{2x}{1 - x^2} \right) = 2\tan^{-1} x \).
2) Since \( x > 0 \), we have \( \tan^{-1}\left( \frac{1}{x} \right) = \cot^{-1} x \).
Substituting these simplifications back into the expression:
\( y = \frac{1}{2} \left[ 2\tan^{-1} x + 2\cot^{-1} x \right] = \frac{1}{2} \cdot 2 \left[ \tan^{-1} x + \cot^{-1} x \right] = \tan^{-1} x + \cot^{-1} x \).
Using the complementary angle identity:
\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \).
Thus, we get \( y = \frac{\pi}{2} \).
Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) = 0 \).
In simple words: Simplify the inverse tangent terms using identities. The equation reduces to a constant value of \( \pi/2 \), whose derivative is simply 0.

Exam Tip: Always look for inverse trigonometric simplifications first; they can transform a complicated differentiation problem into a simple constant derivative.

Question 17. If \( y = \sin\left[2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right] \) then \( \frac{dy}{dx} \) = ?
Answer:
Let \( x = \cos\theta \implies \theta = \cos^{-1} x \).
Substitute this into the square root expression:
\( \frac{1 - x}{1 + x} = \frac{1 - \cos\theta}{1 + \cos\theta} = \frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)} = \tan^2(\theta/2) \).
Since \( x \in (-1, 1) \), we have \( \theta \in (0, \pi) \implies \theta/2 \in (0, \pi/2) \), which means:
\( \sqrt{\frac{1 - x}{1 + x}} = \tan(\theta/2) \).
Substituting this back into \( y \):
\( y = \sin \left[ 2\tan^{-1}(\tan(\theta/2)) \right] = \sin \left[ 2 \cdot \frac{\theta}{2} \right] = \sin\theta \).
Since \( \cos\theta = x \), we can write \( \sin\theta = \sqrt{1 - x^2} \).
Thus, \( y = \sqrt{1 - x^2} \).
Now differentiate with respect to \( x \) using the chain rule:
\( \frac{dy}{dx} = \frac{d}{dx} (1 - x^2)^{1/2} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \).
In simple words: Substitute \( x = \cos\theta \) to simplify the inner fraction into a tangent term, cancel the inverse functions to get \( y = \sin\theta \), convert it back to \( x \), and differentiate.

Exam Tip: Trigonometric substitutions are highly efficient for simplifying expressions containing terms like \( \sqrt{\frac{1-x}{1+x}} \).

Question 18. If \( 5^x + 5^y = 5^{x+y} \) then prove that \( \frac{dy}{dx} + 5^{y-x} = 0 \).
Answer:
We differentiate both sides of the given equation with respect to \( x \):
\( \frac{d}{dx}(5^x) + \frac{d}{dx}(5^y) = \frac{d}{dx}(5^{x+y}) \)
\( 5^x \ln 5 + 5^y \ln 5 \frac{dy}{dx} = 5^{x+y} \ln 5 \left(1 + \frac{dy}{dx}\right) \).
Dividing both sides by the scalar term \( \ln 5 \):
\( 5^x + 5^y \frac{dy}{dx} = 5^{x+y} + 5^{x+y} \frac{dy}{dx} \).
Substitute \( 5^{x+y} = 5^x + 5^y \) from the original equation into the right-hand side:
\( 5^x + 5^y \frac{dy}{dx} = (5^x + 5^y) + (5^x + 5^y) \frac{dy}{dx} \)
\( 5^x + 5^y \frac{dy}{dx} = 5^x + 5^y + 5^x \frac{dy}{dx} + 5^y \frac{dy}{dx} \).
Subtract common terms from both sides:
\( 0 = 5^y + 5^x \frac{dy}{dx} \)

\( \implies 5^x \frac{dy}{dx} = -5^y \)

\( \implies \frac{dy}{dx} = -\frac{5^y}{5^x} = -5^{y-x} \)

\( \implies \frac{dy}{dx} + 5^{y-x} = 0 \).
Hence proved.
In simple words: Differentiate both sides, divide by \( \ln 5 \), and substitute the original equation value back in to cancel the extra terms and prove the relation.

Exam Tip: Substituting the original equation value back into the differentiated expression is a standard and elegant way to simplify implicit derivative proofs.

Question 19. If \( x\sqrt{1-y^2} + y\sqrt{1-x^2} = a \) then show that \( \frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}} \).
Answer:
Let \( x = \sin u \implies u = \sin^{-1} x \) and \( y = \sin v \implies v = \sin^{-1} y \).
Substituting these into the given equation:
\( \sin u \sqrt{1 - \sin^2 v} + \sin v \sqrt{1 - \sin^2 u} = a \)
\( \sin u \cos v + \sin v \cos u = a \).
Using the trigonometric addition formula:
\( \sin(u + v) = a \implies u + v = \sin^{-1} a \).
Substituting the inverse trigonometric functions back:
\( \sin^{-1} x + \sin^{-1} y = \sin^{-1} a \).
Differentiating both sides with respect to \( x \):
\( \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \)

\( \implies \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \)

\( \implies \frac{dy}{dx} = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = -\sqrt{\frac{1-y^2}{1-x^2}} \).
Hence proved.
In simple words: Substitute sine variables to simplify the equation into a basic addition identity. Differentiating this simplified sum directly yields the desired derivative.

Exam Tip: Using substitution makes this proof much cleaner than trying to differentiate the original nested radical equation directly.

Question 20. If \( \sqrt{1-x^2} + \sqrt{1-y^2} = a(x - y) \) then show that \( \frac{dy}{dx} = \sqrt{\frac{1-y^2}{1-x^2}} \).
Answer:
Let \( x = \sin u \) and \( y = \sin v \).
Substituting these into the equation:
\( \cos u + \cos v = a(\sin u - \sin v) \).
Applying trig sum-to-product formulas:
\( 2\cos\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right) = a \left[ 2\cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right) \right] \).
Dividing both sides by the common non-zero term \( 2\cos\left(\frac{u+v}{2}\right) \):
\( \cos\left(\frac{u-v}{2}\right) = a \sin\left(\frac{u-v}{2}\right) \)

\( \implies \cot\left(\frac{u-v}{2}\right) = a \)

\( \implies \frac{u-v}{2} = \cot^{-1} a \)

\( \implies u - v = 2\cot^{-1} a \).
Substitute back \( u = \sin^{-1} x \) and \( v = \sin^{-1} y \):
\( \sin^{-1} x - \sin^{-1} y = 2\cot^{-1} a \).
Differentiating both sides with respect to \( x \):
\( \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \)

\( \implies \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \)

\( \implies \frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = \sqrt{\frac{1-y^2}{1-x^2}} \).
Hence proved.
In simple words: Substitute sine terms, simplify using trig identity products, convert back to inverse sines, and differentiate both sides to find the positive radical fraction.

Exam Tip: Be sure to write out the sum-to-product trigonometric formulas used in step 2 to show complete working to the examiner.

Question 21. If \( (x + y)^{m+n} = x^m \cdot y^n \) then prove that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer:
Taking the natural logarithm on both sides of the given equation:
\( (m + n) \ln(x + y) = m \ln x + n \ln y \).
Differentiating both sides with respect to \( x \):
\( \frac{m+n}{x+y} \left( 1 + \frac{dy}{dx} \right) = \frac{m}{x} + \frac{n}{y} \frac{dy}{dx} \)
\( \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx} = \frac{m}{x} + \frac{n}{y} \frac{dy}{dx} \).
Group the terms containing \( \frac{dy}{dx} \) on the left side:
\( \left( \frac{m+n}{x+y} - \frac{n}{y} \right) \frac{dy}{dx} = \frac{m}{x} - \frac{m+n}{x+y} \)
\( \left[ \frac{y(m+n) - n(x+y)}{y(x+y)} \right] \frac{dy}{dx} = \frac{m(x+y) - x(m+n)}{x(x+y)} \)
\( \left[ \frac{my + ny - nx - ny}{y(x+y)} \right] \frac{dy}{dx} = \frac{mx + my - mx - nx}{x(x+y)} \)
\( \left[ \frac{my - nx}{y(x+y)} \right] \frac{dy}{dx} = \frac{my - nx}{x(x+y)} \).
Assuming \( my - nx \ne 0 \) and \( x+y \ne 0 \), cancel the common terms:
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \implies \frac{dy}{dx} = \frac{y}{x} \).
Hence proved.
In simple words: Take the natural log of both sides to separate the exponent powers, differentiate implicitly, group the derivative terms, and simplify to find \( y/x \).

Exam Tip: Applying logarithmic differentiation is the best way to handle complex product-power equations of the form \( u^p \cdot v^q \).

Question 22. Find the derivative of \( \tan^{-1}\left( \frac{2x}{1 - x^2} \right) \) w.r.t. \( \sin^{-1}\left( \frac{2x}{1 + x^2} \right) \).
Answer:
Let \( u = \tan^{-1}\left( \frac{2x}{1 - x^2} \right) \) and \( v = \sin^{-1}\left( \frac{2x}{1 + x^2} \right) \).
Using standard inverse trigonometric identities for \( |x| < 1 \):
\( u = 2\tan^{-1} x \)
\( v = 2\tan^{-1} x \).
The derivative of \( u \) with respect to \( v \) is:
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}} = 1 \xb \).
In simple words: Since both functions simplify to the exact same expression \( 2\tan^{-1}x \), the derivative of one with respect to the other is simply 1.

Exam Tip: Recognizing that both formulas are identical representations of \( 2\tan^{-1}x \) saves you from doing long quotient-rule differentiations.

Question 23. Find the derivative of loge(sin x) w.r.t. loga(cos x).
Answer:
Let \( u = \log_e(\sin x) \) and \( v = \log_a(\cos x) = \frac{\log_e(\cos x)}{\log_e a} \).
Differentiating both functions with respect to \( x \):
\( \frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \).
\( \frac{dv}{dx} = \frac{1}{\log_e a} \cdot \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\tan x}{\log_e a} \).
Now compute the parametric derivative \( \frac{du}{dv} \):
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\cot x}{-\frac{\tan x}{\log_e a}} = -\cot x \cdot \frac{\log_e a}{\tan x} = -\cot^2 x \log_e a \).
In simple words: Convert the base of the second log to \( e \), find the individual derivatives with respect to \( x \), and divide them to find \( -\cot^2 x \log_e a \).

Exam Tip: Always use the base change formula \( \log_a b = \frac{\ln b}{\ln a} \) first when dealing with derivatives of non-e bases.

Question 24. If \( x^y + y^x + x^x = m^n \), then find the value of \( \frac{dy}{dx} \).
Answer:
Let \( u = x^y \), \( v = y^x \), and \( w = x^x \).
The given equation becomes \( u + v + w = m^n \).
Differentiating both sides with respect to \( x \):
\( \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0 \) --- (1)
Now, let us calculate the individual derivatives:
1) For \( u = x^y \):
\( \ln u = y \ln x \implies \frac{1}{u}\frac{du}{dx} = \frac{dy}{dx}\ln x + \frac{y}{x} \implies \frac{du}{dx} = x^y \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right) \).
2) For \( v = y^x \):
\( \ln v = x \ln y \implies \frac{1}{v}\frac{dv}{dx} = \ln y + \frac{x}{y}\frac{dy}{dx} \implies \frac{dv}{dx} = y^x \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) \).
3) For \( w = x^x \):
\( \ln w = x \ln x \implies \frac{1}{w}\frac{dw}{dx} = \ln x + 1 \implies \frac{dw}{dx} = x^x (\ln x + 1) \).
Substitute these derivatives back into equation (1):
\( x^y \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right) + y^x \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) + x^x (\ln x + 1) = 0 \).
Rearranging terms to solve for \( \frac{dy}{dx} \):
\( \left( x^y \ln x + x y^{x-1} \right) \frac{dy}{dx} = -\left( y x^{y-1} + y^x \ln y + x^x (\ln x + 1) \right) \)

\( \implies \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \ln y + x^x (\ln x + 1)}{x^y \ln x + x y^{x-1}} \).
In simple words: Represent each variable-base term as an individual function, use logarithmic differentiation to find their derivatives, and group the terms to solve for \( \frac{dy}{dx} \).

Exam Tip: Never take logs directly of a sum like \( \log(A+B) \). You must define separate functions \( u, v, w \) and sum their individual derivatives.

Question 25. If \( x = a \cos^3\theta \), \( y = a \sin^3\theta \) then find \( \frac{d^2y}{dx^2} \).
Answer:
Differentiate the parametric equations with respect to \( \theta \):
\( \frac{dx}{d\theta} = 3a \cos^2\theta (-\sin\theta) = -3a \cos^2\theta \sin\theta \).
\( \frac{dy}{d\theta} = 3a \sin^2\theta \cos\theta \).
Now find the first derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2\theta \cos\theta}{-3a \cos^2\theta \sin\theta} = -\tan\theta \).
Now find the second derivative \( \frac{d^2 y}{dx^2} \) using the chain rule:
\( \frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}(-\tan\theta) \cdot \frac{d\theta}{dx} \)
\( = -\sec^2\theta \cdot \frac{1}{\frac{dx}{d\theta}} \)
\( = -\sec^2\theta \cdot \frac{1}{-3a \cos^2\theta \sin\theta} = \frac{\sec^4\theta \csc\theta}{3a} \).
In simple words: Find \( dy/dx \) by dividing the individual derivatives with respect to \( \theta \). Differentiate the resulting \( -\tan\theta \) with respect to \( \theta \) and multiply by \( \frac{d\theta}{dx} \) to get the final answer.

Exam Tip: The most common error in parametric second derivatives is forgetting to multiply by \( \frac{d\theta}{dx} \). Always include this term to avoid losing marks.

Very Short Answer Type Questions (1 Mark)

Question 26. If \( x = a e^t(\sin t - \cos t) \), \( y = a e^t(\sin t + \cos t) \) then show that \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \) is 1.
Answer:
Differentiating the parametric equations with respect to \( t \):
\( \frac{dx}{dt} = a e^t (\sin t - \cos t) + a e^t (\cos t + \sin t) \)
\( = a e^t (\sin t - \cos t + \cos t + \sin t) = 2 a e^t \sin t \).

\( \frac{dy}{dt} = a e^t (\sin t + \cos t) + a e^t (\cos t - \sin t) \)
\( = a e^t (\sin t + \cos t + \cos t - \sin t) = 2 a e^t \cos t \).

Now compute the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 a e^t \cos t}{2 a e^t \sin t} = \cot t \).

At \( t = \frac{\pi}{4} \):
\( \frac{dy}{dx} = \cot\left(\frac{\pi}{4}\right) = 1 \).
Hence shown.
In simple words: First find the derivatives of \( x \) and \( y \) with respect to \( t \). Divide them to get \( \cot t \), and substitute \( t = \pi/4 \) to prove that the derivative is equal to 1.

Exam Tip: Parametric differentiation requires that you differentiate \( x \) and \( y \) separately before dividing them. Always write out \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) as an intermediate formula to earn full step marks.

Question 27. If \( y = \sin^{-1}\left[ x\sqrt{1-x} - \sqrt{x}\sqrt{1-x^2} \right] \) then find \( \frac{dy}{dx} \).
Answer:
Let \( x = \sin u \implies u = \sin^{-1} x \) and \( \sqrt{x} = \sin v \implies v = \sin^{-1} \sqrt{x} \).
Substitute these variables into the expression for \( y \):
\( y = \sin^{-1}\left[ \sin u \sqrt{1 - \sin^2 v} - \sin v \sqrt{1 - \sin^2 u} \right] \)
\( = \sin^{-1}\left[ \sin u \cos v - \sin v \cos u \right] \).
Using the trigonometric angle subtraction identity:
\( y = \sin^{-1}\left[ \sin(u - v) \right] = u - v \).
Substituting the inverse trigonometric functions back:
\( y = \sin^{-1} x - \sin^{-1} \sqrt{x} \).
Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\sin^{-1} x) - \frac{d}{dx} (\sin^{-1} \sqrt{x}) \)
\( = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) \)
\( = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} \)
\( = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{2\sqrt{x}\sqrt{1-x}} \).
In simple words: Use substitution to turn the complicated inner term into a basic sine subtraction formula. Once simplified, differentiate each part individually.

Exam Tip: Be sure to use the chain rule when differentiating \( \sin^{-1}\sqrt{x} \), remembering to multiply by the derivative of \( \sqrt{x} \), which is \( \frac{1}{2\sqrt{x}} \).

Question 28. If \( y = x^{\log_e x} + (\log_e x)^x \) then find \( \frac{dy}{dx} \).
Answer:
Let \( u = x^{\log x} \) and \( v = (\log x)^x \), so \( y = u + v \).
We differentiate \( u \) and \( v \) separately using logarithmic differentiation:
For \( u = x^{\log x} \):
\( \ln u = \ln\left(x^{\log x}\right) = (\log x)^2 \).
Differentiating with respect to \( x \):
\( \frac{1}{u}\frac{du}{dx} = 2(\log x) \cdot \frac{1}{x} \)

\( \implies \frac{du}{dx} = u \left( \frac{2\log x}{x} \right) = x^{\log x} \left( \frac{2\log x}{x} \right) = 2(\log x) x^{\log x - 1} \).

For \( v = (\log x)^x \):
\( \ln v = \ln\left((\log x)^x\right) = x \ln(\log x) \).
Differentiating with respect to \( x \) using the product rule:
\( \frac{1}{v}\frac{dv}{dx} = 1 \cdot \ln(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \ln(\log x) + \frac{1}{\log x} \)

\( \implies \frac{dv}{dx} = (\log x)^x \left[ \ln(\log x) + \frac{1}{\log x} \right] \).

Since \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \), we have:
\( \frac{dy}{dx} = 2(\log x) x^{\log x - 1} + (\log x)^x \left[ \ln(\log x) + \frac{1}{\log x} \right] \).
In simple words: Since we are differentiating variable bases raised to variable powers, define them as separate functions \( u \) and \( v \), apply logarithms to solve, and add the derivatives together.

Exam Tip: Never take logs directly across addition terms like \( \log(A+B) \). Split the function into separate parts \( u \) and \( v \) before using logarithms.

Question 29. Differentiate \( x^{x^x} \) w.r.t. x.
Answer:
Let \( y = x^{x^x} \).
Taking the natural logarithm on both sides:
\( \ln y = x^x \ln x \).
Taking the logarithm again on both sides:
\( \ln(\ln y) = \ln\left(x^x \ln x\right) = \ln(x^x) + \ln(\ln x) = x \ln x + \ln(\ln x) \).
Now differentiate both sides with respect to \( x \):
\( \frac{1}{\ln y} \cdot \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x) + \frac{d}{dx}(\ln(\ln x)) \)
\( \frac{1}{y \ln y} \frac{dy}{dx} = (1 \cdot \ln x + x \cdot \frac{1}{x}) + \frac{1}{\ln x} \cdot \frac{1}{x} = \ln x + 1 + \frac{1}{x \ln x} \)

\( \implies \frac{dy}{dx} = y \ln y \left[ \ln x + 1 + \frac{1}{x \ln x} \right] \).
Substitute back \( y = x^{x^x} \) and \( \ln y = x^x \ln x \):
\( \frac{dy}{dx} = x^{x^x} \cdot (x^x \ln x) \left[ \ln x + 1 + \frac{1}{x \ln x} \right] \)
\( = x^{x^x + x} \ln x \left[ \ln x + 1 + \frac{1}{x \ln x} \right] \).
In simple words: Apply logarithms twice to fully simplify the stacked powers, then differentiate implicitly and substitute back the original values to solve.

Exam Tip: Double logarithmic differentiation is the most reliable way to differentiate tower functions of the form \( f(x)^{g(x)^{h(x)}} \).

Question 30. Find \( \frac{dy}{dx} \), if \( (\cos x)^y = (\cos y)^x \).
Answer:
Taking the natural logarithm on both sides of the given equation:
\( y \ln(\cos x) = x \ln(\cos y) \).
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{dy}{dx} \ln(\cos x) + y \cdot \frac{1}{\cos x} (-\sin x) = 1 \cdot \ln(\cos y) + x \cdot \frac{1}{\cos y} (-\sin y) \frac{dy}{dx} \)
\( \frac{dy}{dx} \ln(\cos x) - y \tan x = \ln(\cos y) - x \tan y \frac{dy}{dx} \).
Grouping the \( \frac{dy}{dx} \) terms on the left side:
\( \frac{dy}{dx} \ln(\cos x) + x \tan y \frac{dy}{dx} = \ln(\cos y) + y \tan x \)
\( \frac{dy}{dx} \left[ \ln(\cos x) + x \tan y \right] = \ln(\cos y) + y \tan x \)

\( \implies \frac{dy}{dx} = \frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y} \).
In simple words: Take the natural log of both sides, differentiate implicitly using product and chain rules, and group the derivative terms to solve.

Exam Tip: Remember to apply the chain rule when differentiating \( \ln(\cos y) \), which introduces both \( -\tan y \) and \( \frac{dy}{dx} \).

Question 31. If \( y = \tan^{-1}\left( \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \right) \), where \( \frac{\pi}{2} < x < \pi \), find \( \frac{dy}{dx} \).
Answer:
Using trigonometric half-angle formulas, we know that:
\( 1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2 \)
\( 1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2 \).
Since \( \frac{\pi}{2} < x < \pi \implies \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \).
In this quadrant, \( \sin(x/2) > \cos(x/2) \), which means that:
\( \sqrt{1 - \sin x} = |\cos(x/2) - \sin(x/2)| = \sin(x/2) - \cos(x/2) \).
Also, \( \sqrt{1 + \sin x} = \cos(x/2) + \sin(x/2) \).
Substituting these roots back into the fraction:
\( \sqrt{1 + \sin x} - \sqrt{1 - \sin x} = (\cos(x/2) + \sin(x/2)) - (\sin(x/2) - \cos(x/2)) = 2\cos(x/2) \).
\( \sqrt{1 + \sin x} + \sqrt{1 - \sin x} = (\cos(x/2) + \sin(x/2)) + (\sin(x/2) - \cos(x/2)) = 2\sin(x/2) \).
So, the argument simplifies to:
\( \frac{\sqrt{1+\sin x} - \sqrt{1-\sin x}}{\sqrt{1+\sin x} + \sqrt{1-\sin x}} = \frac{2\cos(x/2)}{2\sin(x/2)} = \cot(x/2) = \tan\left(\frac{\pi}{2} - \frac{x}{2}\right) \).
Thus:
\( y = \tan^{-1}\left[\tan\left(\frac{\pi}{2} - \frac{x}{2}\right)\right] = \frac{\pi}{2} - \frac{x}{2} \).
Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - \frac{x}{2} \right) = -\frac{1}{2} \).
In simple words: Rewrite the square root terms using half-angle trig identities. Be careful with the quadrant interval, simplify the fraction to a single tangent term, and differentiate.

Exam Tip: Since \( x/2 \) is in the second quadrant where sine is larger than cosine, the term \( \sqrt{1-\sin x} \) must simplify to \( \sin(x/2) - \cos(x/2) \) instead of \( \cos(x/2) - \sin(x/2) \).

Question 32. If \( x = \sin\left(\frac{1}{a} \log_e y\right) \) then show that \( (1 - x^2) y'' - xy' - a^2 y = 0 \).
Answer:
We rewrite the given equation:
\( \sin^{-1} x = \frac{1}{a} \log_e y \)

\( \implies \log_e y = a \sin^{-1} x \)

\( \implies y = e^{a \sin^{-1} x} \).
Differentiating with respect to \( x \):
\( y' = e^{a \sin^{-1} x} \cdot \frac{a}{\sqrt{1 - x^2}} = \frac{a y}{\sqrt{1 - x^2}} \).
Squaring both sides of the equation:
\( y'^2 (1 - x^2) = a^2 y^2 \).
Differentiating both sides with respect to \( x \):
\( 2y' y'' (1 - x^2) + y'^2 (-2x) = a^2 (2y y') \).
Since \( y' \ne 0 \), dividing both sides by the common factor \( 2y' \):
\( y'' (1 - x^2) - x y' = a^2 y \)

\( \implies (1 - x^2)y'' - xy' - a^2 y = 0 \).
Hence shown.
In simple words: Rearrange the equation to express \( y \) as an exponent, find the first derivative, square both sides to remove the square root, and differentiate again to prove the differential equation.

Exam Tip: Squaring both sides after finding the first derivative is a standard mathematical trick that prevents difficult quotient rule steps in the second derivative.

Question 33. Differentiate \( (\log x)^{\log x} \), \( x > 1 \) w.r.t. x.
Answer:
Let \( y = (\log x)^{\log x} \).
Taking the natural logarithm on both sides:
\( \ln y = \log x \cdot \ln(\log x) \).
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \ln(\log x) + \log x \cdot \frac{d}{dx}[\ln(\log x)] \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \ln(\log x) + \log x \cdot \left[ \frac{1}{\log x} \cdot \frac{1}{x} \right] \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\ln(\log x)}{x} + \frac{1}{x} = \frac{\ln(\log x) + 1}{x} \)

\( \implies \frac{dy}{dx} = y \left[ \frac{\ln(\log x) + 1}{x} \right] \).
Substituting the original value of \( y \) back:
\( \frac{dy}{dx} = (\log x)^{\log x} \left[ \frac{\ln(\log x) + 1}{x} \right] \).
In simple words: Take the natural log of both sides, differentiate using the product and chain rules, and substitute the original function back to find the final derivative.

Exam Tip: Clearly write out the product rule steps. When differentiating \( \ln(\log x) \), remember to apply the chain rule to get \( \frac{1}{x \log x} \).

Question 34. If sin y = x sin (a + y) then show that \( \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \).
Answer:
We can isolate \( x \) in terms of \( y \):
\( x = \frac{\sin y}{\sin(a + y)} \).
Differentiating with respect to \( y \) using the quotient rule:
\( \frac{dx}{dy} = \frac{\cos y \sin(a + y) - \sin y \cos(a + y)}{\sin^2(a + y)} \).
Using the trigonometric identity \( \sin A \cos B - \cos A \sin B = \sin(A - B) \):
\( \frac{dx}{dy} = \frac{\sin(a + y - y)}{\sin^2(a + y)} = \frac{\sin a}{\sin^2(a + y)} \xb \).
Now, taking the reciprocal to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{\sin^2(a + y)}{\sin a} \).
Hence shown.
In simple words: Express \( x \) in terms of \( y \), differentiate with respect to \( y \) using the quotient rule, simplify the numerator using trig identities, and take the reciprocal to get \( dy/dx \).

Exam Tip: Differentiating \( x \) with respect to \( y \) first is much easier than implicit differentiation of \( y \) with respect to \( x \) for this specific proof.

Question 35. If \( y = \sin^{-1} x \), find \( \frac{d^2 y}{dx^2} \) in terms of y.
Answer:
Given \( y = \sin^{-1} x \implies x = \sin y \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\cos y} = \sec y \).
Now differentiate again with respect to \( x \) to find the second derivative:
\( \frac{d^2 y}{dx^2} = \frac{d}{dx}(\sec y) = \sec y \tan y \cdot \frac{dy}{dx} \).
Substituting \( \frac{dy}{dx} = \sec y \) into the equation:
\( \frac{d^2 y}{dx^2} = \sec y \tan y \cdot \sec y = \sec^2 y \tan y \).
In simple words: Write \( x \) as \( \sin y \), find \( dy/dx \) as \( \sec y \), and differentiate again with respect to \( x \) using the chain rule to write the final second derivative in terms of \( y \).

Exam Tip: Always multiply by \( \frac{dy}{dx} \) when differentiating a \( y \)-variable term with respect to \( x \) during second-order derivative steps.

Question 36. If \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), then show that \( \frac{d^2y}{dx^2} = \frac{-b^4}{a^2 y^3} \).
Answer:
Differentiating the given ellipse equation with respect to \( x \):
\( \frac{2x}{a^2} + \frac{2y}{b^2} y' = 0 \implies y' = -\frac{b^2 x}{a^2 y} \).
Differentiating again with respect to \( x \) using the quotient rule:
\( y'' = \frac{d}{dx}\left(-\frac{b^2 x}{a^2 y}\right) = -\frac{b^2}{a^2} \left[ \frac{1 \cdot y - x \cdot y'}{y^2} \right] \).
Substitute \( y' = -\frac{b^2 x}{a^2 y} \) into the derivative:
\( y'' = -\frac{b^2}{a^2 y^2} \left[ y - x \left(-\frac{b^2 x}{a^2 y}\right) \right] = -\frac{b^2}{a^2 y^2} \left[ \frac{a^2 y^2 + b^2 x^2}{a^2 y} \right] \).
Using the original equation, we know that \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies b^2 x^2 + a^2 y^2 = a^2 b^2 \).
Substitute this identity back:
\( y'' = -\frac{b^2}{a^2 y^2} \left[ \frac{a^2 b^2}{a^2 y} \right] = -\frac{b^4}{a^2 y^3} \).
Hence shown.
In simple words: Find the first derivative, use the quotient rule to differentiate again, substitute the first derivative value back in, and simplify using the original ellipse equation to prove the final fractional form.

Exam Tip: Substituting the original equation's identity \( b^2x^2 + a^2y^2 = a^2b^2 \) at the end is a standard step needed to simplify second-order derivative proofs.

Question 37. If \( y = e^{a \cos^{-1} x} \), \( -1 \le x \le 1 \), show that \( (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \).
Answer:
Differentiating the function with respect to \( x \):
\( y' = e^{a \cos^{-1} x} \cdot \left( \frac{-a}{\sqrt{1 - x^2}} \right) = \frac{-a y}{\sqrt{1 - x^2}} \).
Squaring both sides of the equation:
\( y'^2 (1 - x^2) = a^2 y^2 \).
Differentiating both sides with respect to \( x \):
\( 2y' y'' (1 - x^2) + y'^2 (-2x) = a^2 (2y y') \).
Dividing both sides of the equation by the common non-zero factor \( 2y' \):
\( y'' (1 - x^2) - x y' = a^2 y \)

\( \implies (1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \).
Hence shown.
In simple words: Differentiate once, square both sides to eliminate the square root, and differentiate again to solve for the target second-order differential equation.

Exam Tip: Using the squaring trick after the first derivative prevents tedious quotient-rule steps and algebraic errors when finding the second derivative.

Question 38. If \( y^3 = 3ax^2 - x^3 \) then prove that \( \frac{d^2 y}{dx^2} = \frac{-2a^2 x^2}{y^5} \).
Answer:
Differentiating both sides with respect to \( x \):
\( 3y^2 y' = 6ax - 3x^2 \implies y^2 y' = 2ax - x^2 \) --- (1)
Differentiating again with respect to \( x \) using the product rule:
\( \frac{d}{dx}(y^2) \cdot y' + y^2 \cdot y'' = \frac{d}{dx}(2ax - x^2) \)
\( 2y y'^2 + y^2 y'' = 2a - 2x \).
Multiply both sides of the equation by \( y^3 \):
\( 2y^4 y'^2 + y^5 y'' = 2(a - x)y^3 \).
From equation (1), we have \( y^4 y'^2 = (y^2 y')^2 = (2ax - x^2)^2 \). Substitute this value:
\( 2(2ax - x^2)^2 + y^5 y'' = 2(a - x)(3ax^2 - x^3) \)
\( 2(4a^2 x^2 - 4ax^3 + x^4) + y^5 y'' = 2(3a^2 x^2 - ax^3 - 3ax^3 + x^4) \)
\( 8a^2 x^2 - 8ax^3 + 2x^4 + y^5 y'' = 6a^2 x^2 - 8ax^3 + 2x^4 \).
Subtract common terms from both sides of the equation:
\( 8a^2 x^2 + y^5 y'' = 6a^2 x^2 \)
\( y^5 y'' = 6a^2 x^2 - 8a^2 x^2 = -2a^2 x^2 \)

\( \implies y'' = \frac{-2a^2 x^2}{y^5} \).
Hence proved.
In simple words: Find the first derivative, differentiate again, multiply the equation by \( y^3 \) to match powers, substitute algebraic terms, and subtract to find the final derivative formula.

Exam Tip: Scaling the second derivative equation by a power of \( y \) is a helpful technique that eliminates fraction denominators and simplifies polynomial checks.

Question 39. Verify Rolle's theorem for the function, \( y = x^2 + 2 \) in the interval [a, b] where a = –2, b = 2.
Answer:
Let \( f(x) = x^2 + 2 \) on the closed interval \( [-2, 2] \).
We check the three criteria for Rolle's Theorem:
1) \( f(x) \) is a polynomial function, so it is continuous on \( [-2, 2] \).
2) \( f(x) \) is differentiable on \( (-2, 2) \) with \( f'(x) = 2x \).
3) Check boundary values:
\( f(-2) = (-2)^2 + 2 = 6 \) and \( f(2) = 2^2 + 2 = 6 \).
Since \( f(-2) = f(2) \), all three conditions of Rolle's Theorem are satisfied.
Therefore, there must exist at least one real number \( c \in (-2, 2) \) such that:
\( f'(c) = 0 \implies 2c = 0 \implies c = 0 \xb \).
Since \( 0 \in (-2, 2) \), Rolle's Theorem is verified.
In simple words: Verify that the quadratic function is continuous, differentiable, and has equal values at the endpoints. Since all conditions match, find the point \( c \) where the slope is 0, which lies in the interval.

Exam Tip: Always explicitly list all three criteria (continuity, differentiability, and endpoint equality) to show a complete proof of the theorem.

Question 40. Verify Mean Value Theorem for the function, f(x) = \( x^2 \) in [2, 4].
Answer:
We check the criteria of the Mean Value Theorem (MVT) for \( f(x) = x^2 \) on \( [2, 4] \):
1) \( f(x) \) is a polynomial function, so it is continuous on \( [2, 4] \).
2) \( f(x) \) is differentiable on \( (2, 4) \) with \( f'(x) = 2x \).
According to MVT, there exists at least one real number \( c \in (2, 4) \) such that:
\( f'(c) = \frac{f(4) - f(2)}{4 - 2} \).
Evaluate the functional values:
\( f(4) = 4^2 = 16 \) and \( f(2) = 2^2 = 4 \).
Substituting these values into the equation:
\( 2c = \frac{16 - 4}{4 - 2} = \frac{12}{2} = 6 \)

\( \implies c = 3 \).
Since \( 3 \in (2, 4) \), the Mean Value Theorem is verified.
In simple words: Check that the function is continuous and differentiable. Set the derivative slope equal to the secant slope between the endpoints, and solve to find \( c = 3 \), which lies inside the interval.

Exam Tip: Ensure that the solved point \( c \) lies strictly inside the open interval \( (a, b) \), not on the boundaries.

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