CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 01

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 1. The Continuity And Differentiability questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Continuity And Differentiability concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Continuity And Differentiability worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Continuity And Differentiability chapter and other subjects too. Use them for better understanding of the subjects.

Question 1. If \( y = a^x + e^x + x^x + x^a + a^a \). Find \( \frac{dy}{dx} \) at x = a.
Answer:
Let us represent the term \( x^x \) using an auxiliary variable \( u \):
\( y = a^x + e^x + u + x^a + a^a \)
Differentiating each term with respect to \( x \) yields:
\( \frac{dy}{dx} = \frac{d}{dx}(a^x) + \frac{d}{dx}(e^x) + \frac{du}{dx} + \frac{d}{dx}(x^a) + \frac{d}{dx}(a^a) \)
Applying standard differentiation rules:
\( \frac{dy}{dx} = a^x \log a + e^x + \frac{du}{dx} + a x^{a-1} + 0 \) ----(i)
To determine \( \frac{du}{dx} \), we take natural logarithms on both sides of \( u = x^x \):
\( \log u = x \log x />\)
Now, differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{u} \cdot \frac{du}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \)
\( \frac{1}{u} \cdot \frac{du}{dx} = 1 + \log x \)

\( \implies \frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x) \)
Substituting this derivative back into equation (i):
\( \frac{dy}{dx} = a^x \log a + e^x + x^x(1 + \log x) + a x^{a-1} />\)
Finally, we evaluate this derivative at \( x = a \):
\( \left(\frac{dy}{dx}\right)_{x=a} = a^a \log a + e^a + a^a(1 + \log a) + a \cdot a^{a-1} \)
\( \left(\frac{dy}{dx}\right)_{x=a} = a^a \log a + e^a + a^a + a^a \log a + a^a \)
Combining the identical terms gives:
\( \left(\frac{dy}{dx}\right)_{x=a} = e^a + 2a^a(1 + \log a) \)
In simple words: To find the derivative of a sum, differentiate each part separately. For the tricky term \( x^x \), use logarithmic differentiation first before substituting \( x = a \) into the combined final result.

Exam Tip: Always remember that \( a^a \) is a constant, so its derivative is zero. Do not mistakenly apply the power rule or exponential rule to it.

Question 2. If \( y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \). Find \( \frac{dy}{dx} \).
Answer:
We can express the given equation as:
\( y = \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]^{1/2} \)
Taking the natural logarithm of both sides:
\( \log y = \log \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]^{1/2} />\)
Using logarithmic properties to expand the right-hand side:
\( \log y = \frac{1}{2} [ \log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) - \log(x-5) ] />\)
Differentiating both sides with respect to \( x \) using the chain rule:
\( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] />\)
Multiplying both sides by \( y \):
\( \frac{dy}{dx} = \frac{y}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] />\)
Substituting the original value of \( y \) back into the derivative:
\( \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right] \)
In simple words: When dealing with a complex product and quotient under a square root, taking the natural log first simplifies the expression into a sum and difference of simple log terms, making differentiation much easier.

Exam Tip: Keep the term \( y \) factored out until the very end, then substitute it back in its original radical form to write the final derivative clearly.

Question 3. (i) \( y = (x^x)^x \) (ii) \( y = x^{x^x} \). Find \( \frac{dy}{dx} \).
Answer:
(i) For the first expression, we can simplify the exponent first using power rules:
\( y = (x^x)^x \)

\( \implies y = x^{x^2} \)
Taking the natural logarithm of both sides:
\( \log y = x^2 \log x />\)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{y} \cdot \frac{dy}{dx} = x^2 \cdot \frac{1}{x} + \log x \cdot (2x) \)
\( \frac{1}{y} \cdot \frac{dy}{dx} = x + 2x \log x \)
Multiplying both sides by \( y \):
\( \frac{dy}{dx} = y(x + 2x \log x) = x^{x^2} (x + 2x \log x) \)

(ii) For the second expression, we have:
\( y = x^{x^x} \)
Let us substitute the exponent with an auxiliary variable \( u = x^x \):
\( y = x^u \)
Taking natural logarithm on both sides:
\( \log y = u \log x />\)
Differentiating both sides with respect to \( x \) using the product rule:
\( \frac{1}{y} \cdot \frac{dy}{dx} = u \cdot \frac{1}{x} + \log x \cdot \frac{du}{dx} \)
\( \frac{dy}{dx} = y \left( \frac{u}{x} + \log x \cdot \frac{du}{dx} \right) \) ----(ii)
To find \( \frac{du}{dx} \) for \( u = x^x \), we use logarithmic differentiation:
\( \log u = x \log x \)
Differentiating with respect to \( x \):
\( \frac{1}{u} \cdot \frac{du}{dx} = 1 + \log x \)

\( \implies \frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x) \)
Plugging the values of \( u \) and \( \frac{du}{dx} \) back into equation (ii):
\( \frac{dy}{dx} = x^{x^x} \left( \frac{x^x}{x} + \log x \cdot x^x(1 + \log x) \right) \)
Factoring out \( x^x \) gives the final simplified result:
\( \frac{dy}{dx} = x^{x^x} \cdot x^x \left( \frac{1}{x} + \log x(1 + \log x) \right) \)
In simple words: (i) For the first part, simplify the exponent first using power rules, \( (x^a)^b = x^{ab} \), and then use log differentiation. (ii) For the second part, since the exponent itself is a function of \( x \) as well, we treat it as a separate helper function \( u \) and apply the chain rule.

Exam Tip: Be extremely careful with the difference between \( (x^x)^x = x^{x^2} \) and \( x^{x^x} \). They are completely different functions and require different mathematical treatments.

Question 4. Given that \( \cos\frac{x}{2} \cdot \cos\frac{x}{4} \cdot \cos\frac{x}{8} \dots \infty = \frac{\sin x}{x} \) show that \( \frac{1}{2^2} \cdot \sec^2\left(\frac{x}{2}\right) + \frac{1}{2^4} \cdot \sec^2\left(\frac{x}{4}\right) + \dots = \text{cosec}^2 x - \frac{1}{x^2} \)
Answer:
We begin with the given infinite product relationship:
\( \cos\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{4}\right) \cdot \cos\left(\frac{x}{8}\right) \dots \infty = \frac{\sin x}{x} />\)
Applying the natural logarithm to both sides converts the product into an infinite sum:
\( \log\left(\cos\frac{x}{2}\right) + \log\left(\cos\frac{x}{4}\right) + \log\left(\cos\frac{x}{8}\right) + \dots = \log(\sin x) - \log x /> />\)
Differentiating each term with respect to \( x \) using the chain rule:
\( \frac{1}{\cos(x/2)} \cdot \left(-\sin\frac{x}{2}\right) \cdot \frac{1}{2} + \frac{1}{\cos(x/4)} \cdot \left(-\sin\frac{x}{4}\right) \cdot \frac{1}{4} + \dots = \frac{1}{\sin x} \cdot \cos x - \frac{1}{x} />\)
Simplifying the trigonometric terms into tangents:
\( -\frac{1}{2}\tan\left(\frac{x}{2}\right) - \frac{1}{4}\tan\left(\frac{x}{4}\right) - \dots = \cot x - \frac{1}{x} />\)
Multiplying the entire equation by \( -1 \):

\( \implies \frac{1}{2}\tan\left(\frac{x}{2}\right) + \frac{1}{4}\tan\left(\frac{x}{4}\right) + \dots = -\cot x + \frac{1}{x} \)
Differentiating both sides with respect to \( x \) once again:
\( \frac{1}{2}\sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} + \frac{1}{4}\sec^2\left(\frac{x}{4}\right) \cdot \frac{1}{4} + \dots = -(-\text{cosec}^2 x) - \frac{1}{x^2} \)

\( \implies \frac{1}{2^2}\sec^2\left(\frac{x}{2}\right) + \frac{1}{2^4}\sec^2\left(\frac{x}{4}\right) + \dots = \text{cosec}^2 x - \frac{1}{x^2} \)
This proves the required identity.
In simple words: This proof uses logarithmic differentiation to convert a complicated product of cosine terms into a sum of logarithms. Differentiating twice transforms the cosines first into tangents, and then into secants, which matches the right-hand side.

Exam Tip: Pay close attention to the chain rule when differentiating \( \tan(x/2^n) \) - the extra factor of \( 1/2^n \) from the inner derivative creates the \( 1/2^{2n} \) terms in the second derivative.

Question 5. If \( y = \frac{ax^2}{(x-a)(x-b)(x-c)} + \frac{bx}{(x-b)(x-c)} + \frac{c}{(x-c)} + 1 \). Show that \( \frac{dy}{dx} = \frac{y}{x} \left\{ \frac{a}{a-x} + \frac{b}{b-x} + \frac{c}{c-x} \right\} \)
Answer:
To simplify the expression for \( y \) efficiently, we combine the terms starting from the right-hand side. Let's add the last two terms:
\( \frac{c}{x-c} + 1 = \frac{c + x - c}{x-c} = \frac{x}{x-c} />\)
Now, add this result to the second term:
\( \frac{bx}{(x-b)(x-c)} + \frac{x}{x-c} = \frac{bx + x(x-b)}{(x-b)(x-c)} = \frac{bx + x^2 - bx}{(x-b)(x-c)} = \frac{x^2}{(x-b)(x-c)} />\)
Now, add this result to the first term of the expression:
\( \frac{ax^2}{(x-a)(x-b)(x-c)} + \frac{x^2}{(x-b)(x-c)} = \frac{ax^2 + x^2(x-a)}{(x-a)(x-b)(x-c)} = \frac{ax^2 + x^3 - ax^2}{(x-a)(x-b)(x-c)} = \frac{x^3}{(x-a)(x-b)(x-c)} />\)
Therefore, the function simplifies elegantly to:
\( y = \frac{x^3}{(x-a)(x-b)(x-c)} />\)
Taking the natural logarithm of both sides:
\( \log y = 3\log x - \log(x-a) - \log(x-b) - \log(x-c) />\)
Differentiating both sides with respect to \( x \):
\( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{3}{x} - \frac{1}{x-a} - \frac{1}{x-b} - \frac{1}{x-c} \)

\( \implies \frac{dy}{dx} = y \left[ \frac{3}{x} - \frac{1}{x-a} - \frac{1}{x-b} - \frac{1}{x-c} \right] \)
By separating \( \frac{3}{x} \) as \( \frac{1}{x} + \frac{1}{x} + \frac{1}{x} \), we can pair each part with the other fractional terms:
\( \frac{dy}{dx} = y \left[ \left(\frac{1}{x} - \frac{1}{x-a}\right) + \left(\frac{1}{x} - \frac{1}{x-b}\right) + \left(\frac{1}{x} - \frac{1}{x-c}\right) \right] />\)
Simplifying each grouped pair:
\( \frac{1}{x} - \frac{1}{x-a} = \frac{x-a-x}{x(x-a)} = \frac{-a}{x(x-a)} = \frac{a}{x(a-x)} \)
Applying this simplification to all three pairs:
\( \frac{dy}{dx} = y \left[ \frac{a}{x(a-x)} + \frac{b}{x(b-x)} + \frac{c}{x(c-x)} \right] \)

\( \implies \frac{dy}{dx} = \frac{y}{x} \left[ \frac{a}{a-x} + \frac{b}{b-x} + \frac{c}{c-x} \right] \)
This proves the required derivative format.
In simple words: Instead of taking the LCM of all terms at once, we simplify the expression from right to left, combining two terms at a time. This reduces the entire expression to a single fraction, which we differentiate easily using logarithms.

Exam Tip: Splitting \( \frac{3}{x} \) into three separate \( \frac{1}{x} \) terms is the crucial trick to match the required format of the proof without tedious algebraic expansion.

Parametric Functions

Question 6. If \( x = a \sec^3\theta \) and \( y = a \tan^3\theta \). Find \( \frac{dy}{dx} \) at \( \theta = \pi/3 \).
Answer:
We are given the parametric coordinates:
\( x = a \sec^3\theta />\)
Differentiating \( x \) with respect to the parameter \( \theta \):
\( \frac{dx}{d\theta} = a \cdot 3\sec^2\theta \cdot (\sec\theta \tan\theta) = 3a \sec^3\theta \tan\theta />\)
Next, we have:
\( y = a \tan^3\theta />\)
Differentiating \( y \) with respect to \( \theta \):
\( \frac{dy}{d\theta} = a \cdot 3\tan^2\theta \cdot (\sec^2\theta) = 3a \tan^2\theta \sec^2\theta />\)
Using the parametric derivative formula:
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \tan^2\theta \sec^2\theta}{3a \sec^3\theta \tan\theta} \)

\( \implies \frac{dy}{dx} = \sin\theta \)
Evaluating this derivative at \( \theta = \pi/3 \):
\( \left( \frac{dy}{dx} \right)_{\theta=\pi/3} = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)
In simple words: For parametric equations, we find the derivatives of x and y separately with respect to the parameter \( \theta \), then divide them. Simplifying the trigonometric terms gives \( \sin\theta \), which we then evaluate at the given angle.

Exam Tip: Always simplify the trigonometric ratio of \( \frac{dy/d\theta}{dx/d\theta} \) completely before substituting the value of the parameter to make the calculation straightforward.

Question 7. If \( x = a(\cos\theta + \theta \sin\theta) \) and \( y = a(\sin\theta - \theta \cos\theta) \). Find \( \frac{dy}{dx} \) at \( \theta = \pi/6 \).
Answer:
First, we differentiate the parametric term \( x \) with respect to \( \theta \) using the product rule:
\( \frac{dx}{d\theta} = a [-\sin\theta + (\theta \cos\theta + \sin\theta \cdot 1)] \)

\( \implies \frac{dx}{d\theta} = a\theta\cos\theta \)
Next, we differentiate the term \( y \) with respect to \( \theta \) using the product rule:
\( \frac{dy}{d\theta} = a [\cos\theta - (\theta (-\sin\theta) + \cos\theta \cdot 1)] \)
\( \frac{dy}{d\theta} = a [\cos\theta + \theta \sin\theta - \cos\theta] \)

\( \implies \frac{dy}{d\theta} = a\theta\sin\theta \)
Now, calculating the derivative of \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \theta \sin\theta}{a \theta \cos\theta} \)

\( \implies \frac{dy}{dx} = tan\theta \)
Now we evaluate this at \( \theta = \pi/6 \):
\( \left( \frac{dy}{dx} \right)_{\theta=\pi/6} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \)
In simple words: Using the product rule, we find the derivatives of both x and y. Dividing them simplifies elegantly to \( \tan\theta \), which we evaluate at \( \pi/6 \) to get \( 1/\sqrt{3} \).

Exam Tip: Watch the negative signs carefully when applying the product rule to \( \theta \cos\theta \); a small sign error here is a very common trap.

Question 8. If \( x = cos^{-1} \left(\frac{1}{\sqrt{1+t^2}}\right) \) and \( y = sin^{-1} \left(\frac{t}{\sqrt{1+t^2}}\right) \). Find \( \frac{dy}{dx} \).
Answer:
To avoid tedious direct differentiation, we use the trigonometric substitution \( t = \tan\theta \) (which means \( \theta = \tan^{-1} t \)).
For \( x \):
\( x = \cos^{-1}\left(\frac{1}{\sqrt{1+\tan^2\theta}}\right) = \cos^{-1}\left(\frac{1}{\sec\theta}\right) \)

\( \implies x = \cos^{-1}(\cos\theta) = \theta \)
Substituting \( \theta \) back in terms of \( t \) gives:
\( x = \tan^{-1} t />\)
For \( y \):
\( y = \sin^{-1}\left(\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}\right) = \sin^{-1}\left(\frac{\tan\theta}{\sec\theta}\right) \)

\( \implies y = \sin^{-1}(\sin\theta) = \theta \)
Substituting \( \theta \) back in terms of \( t \) gives:
\( y = \tan^{-1} t />\)
Differentiating both \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{1}{1+t^2} />\)
\( \frac{dy}{dt} = \frac{1}{1+t^2} />\)
Now, calculating \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{1+t^2}}{\frac{1}{1+t^2}} = 1 \)
In simple words: Instead of differentiating directly, we use trigonometric substitution \( t = \tan\theta \). This simplifies both x and y to simply \( \tan^{-1} t \). Since both functions are identical, their derivative with respect to each other is simply 1.

Exam Tip: Using substitution for inverse trigonometric functions is much faster and less prone to errors than applying the direct chain rule.

Question 9. If \( x = a \left(\cos t + \log\left(\tan \frac{t}{2}\right)\right) \) and \( y = a \sin t \). Find \( \frac{dy}{dx} \).
Answer:
First, we differentiate \( x \) with respect to \( t \):
\( \frac{dx}{dt} = a \left[ -\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} \right] />\)
Let us simplify the trigonometric term inside the brackets:
\( \frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} = \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2} = \frac{1}{2\sin(t/2)\cos(t/2)} = \frac{1}{\sin t} />\)
Substituting this back into \( \frac{dx}{dt} \):
\( \frac{dx}{dt} = a \left[ -\sin t + \frac{1}{\sin t} \right] = a \left[ \frac{1 - \sin^2 t}{\sin t} \right] = a \frac{\cos^2 t}{\sin t} /> />\)
Next, we differentiate \( y \) with respect to \( t \):
\( \frac{dy}{dt} = a \cos t />\)
Using the formula for parametric derivatives:
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} = \frac{\sin t}{\cos t} = \tan t \)
In simple words: Differentiating the logarithmic tangent term simplifies nicely to \( \frac{1}{\sin t} \) using double-angle identities. Combining this with \( -\sin t \) gives \( \frac{\cos^2 t}{\sin t} \). Dividing the derivative of y by this result leaves us with \( \tan t \).

Exam Tip: This is a classic board exam question. Ensure you write out the identity \( 2 \sin(t/2) \cos(t/2) = \sin t \) clearly in your steps, as examiners specifically look for it.

Question 10. If \( x = \sqrt{a^{\sin^{-1} t}} \) and \( y = \sqrt{a^{\cos^{-1} t}} \). Show that \( \frac{dy}{dx} = -\frac{y}{x} \).
Answer:
Rather than differentiating both parametric equations separately, we can solve this problem in a much simpler way by multiplying the two expressions:
\( x \cdot y = \sqrt{a^{\sin^{-1} t}} \cdot \sqrt{a^{\cos^{-1} t}} />\)
\( x \cdot y = \sqrt{a^{\sin^{-1} t + \cos^{-1} t}} />\)
We know the standard inverse trigonometric identity:
\( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} /> />\)
Substituting this into our equation:
\( x \cdot y = \sqrt{a^{\pi/2}} />\)
Since the right-hand side is a constant, differentiating both sides with respect to \( x \) using the product rule gives:
\( x \cdot \frac{dy}{dx} + y \cdot 1 = 0 \)

\( \implies \frac{dy}{dx} = -\frac{y}{x} \)
This proves the required result elegantly.
In simple words: Since \( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \), multiplying x and y simplifies the expression to a constant: \( x \cdot y = \sqrt{a^{\pi/2}} \). Differentiating this simple product directly yields \( \frac{dy}{dx} = -\frac{y}{x} \) without any tedious calculations.

Exam Tip: Using the identity \( \sin^{-1} t + \cos^{-1} t = \pi/2 \) to turn the system into \( xy = C \) is the ultimate shortcut for this question. It saves time and prevents any algebraic mistakes.

Please click the link below to download CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 1