ML Aggarwal Class 12 Maths Solutions Section A Chapter 06 Differentiation

6.1 Recap

In Class XI, you studied real functions, their limits, continuity, and derivatives. The following key results and examples will help refresh your understanding. (Hopefully, it will be sweet!)

1. Derivative at any point

A function f is said to have a derivative at any point x if it is defined in some (undeleted) neighbourhood of the point x and
\( \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \) exists (finitely).

The value of this limit is called the derivative of f at any point x and is denoted by f'(x), that is,
\( f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \).

Other Notations:

If the function f is written as y = f(x), then its derivative is written as
\( \frac{dy}{dx} \) (or y₁) and so
\( \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \) ...(i)

But y = f(x) \( \implies \) y + δy = f(x + δx)

\( \therefore \) δy = f(x + δx) - f(x) \( \implies \)
\( \frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x} \)

\( \implies \)
\( \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} = \frac{dy}{dx} \) (using (i))

\( \therefore \)
\( \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\delta y}{\delta x} \)

2. \( \frac{d}{dx}(x^n) = nx^{n-1} \), n is a rational number.

3. \( \frac{d}{dx}((ax + b)^n) = n(ax + b)^{n-1} \cdot a \), n is a rational number.

4. The derivative of a constant is zero.

5. If f is a differentiable function at x and g is the function defined by g(x) = cf(x), then g'(x) = cf'(x) where c is a fixed real number.

6. If f and g are differentiable functions at x and if h is the function defined by h(x) = f(x) + g(x), then h'(x) = f'(x) + g'(x).

7. If f and g are differentiable functions at x and if h is the function defined by h(x) = f(x) g(x), then h'(x) = f(x) g'(x) + g(x) f'(x)

That is,
\( \frac{d}{dx}(f(x) \cdot g(x)) = f(x) \cdot \frac{d}{dx}(g(x)) + g(x) \cdot \frac{d}{dx}(f(x)) \)

Thus, the derivative of the product of two differentiable functions = first function × derivative of second function + second function × derivative of first function.

This is known as the product rule.

Extension of the product rule:

If f, g, and h are three differentiable functions at x, then
\( \frac{d}{dx}(f(x) \cdot g(x) \cdot h(x)) = f(x) \cdot g(x) \cdot \frac{d}{dx}(h(x)) + f(x) \cdot h(x) \cdot \frac{d}{dx}(g(x)) + g(x) \cdot h(x) \cdot \frac{d}{dx}(f(x)) \)

8. If f is a differentiable function at x and f(x) ≠ 0, and g is a function defined by \( g(x) = \frac{1}{f(x)} \), then
\( g'(x) = -\frac{f'(x)}{(f(x))^2} \)

That is,
\( \frac{d}{dx}\left(\frac{1}{f(x)}\right) = -\frac{f'(x)}{(f(x))^2} \)

9. If f and g are differentiable functions at x and h is the function defined by \( h(x) = \frac{f(x)}{g(x)} \), g(x) ≠ 0, then
\( h'(x) = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2} \)

That is,
\( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x) \cdot \frac{d}{dx}(f(x)) - f(x) \cdot \frac{d}{dx}(g(x))}{(g(x))^2} \), g(x) ≠ 0

Thus, the derivative of the quotient of two differentiable functions
= \( \frac{\text{deno.} \times \text{derivative of num.} - \text{num.} \times \text{derivative of deno.}}{(\text{denominator})^2} \)

This is known as the quotient rule.

Illustrative Examples

Example 1. Differentiate the following functions:
(i) (2x³ - 7)(9x⁵ + 2x² - 3)
(ii) x(2x - 3)√(x - 2)
(iii) (2x + 1)²(3x - 2)³(4x + 5)⁴

Solution. (i) Let y = (2x³ - 7)(9x⁵ + 2x² - 3). Differentiating with respect to x, we get
\( \frac{dy}{dx} = (2x^3 - 7) \cdot \frac{d}{dx}(9x^5 + 2x^2 - 3) + (9x^5 + 2x^2 - 3) \cdot \frac{d}{dx}(2x^3 - 7) \) (using product rule)

= (2x³ - 7)(45x⁴ + 4x) + (9x⁵ + 2x² - 3)(6x²)

= 90x⁷ + 8x⁴ - 315x⁴ - 28x + 54x⁷ + 12x⁴ - 18x²

= 144x⁷ - 295x⁴ - 18x² - 28x

(ii) Let y = x(2x - 3)√(x - 2) = (2x² - 3x)(x - 2)^(1/2). Differentiating with respect to x, we get
\( \frac{dy}{dx} = (2x^2 - 3x) \cdot \frac{1}{2}(x - 2)^{-1/2} \cdot 1 + (x - 2)^{1/2} \cdot (4x - 3) \) (using product rule)

= \( \frac{2x^2 - 3x}{2\sqrt{x - 2}} + (4x - 3)\sqrt{x - 2} \)

= \( \frac{2x^2 - 3x + 2(4x - 3)(x - 2)}{2\sqrt{x - 2}} \)

= \( \frac{2x^2 - 3x + 2(4x^2 - 11x + 6)}{2\sqrt{x - 2}} \)

= \( \frac{10x^2 - 25x + 12}{2\sqrt{x - 2}} \)

(iii) Let y = (2x + 1)²(3x - 2)³(4x + 5)⁴. Differentiating with respect to x, we get
\( \frac{dy}{dx} = (2x + 1)^2(3x - 2)^3 \cdot \frac{d}{dx}((4x + 5)^4) + (2x + 1)^2(4x + 5)^4 \cdot \frac{d}{dx}((3x - 2)^3) + (3x - 2)^3(4x + 5)^4 \cdot \frac{d}{dx}((2x + 1)^2) \) (using extension of product rule)

= (2x + 1)²(3x - 2)³ \cdot 4(4x + 5)³ \cdot 4 + (2x + 1)²(4x + 5)⁴ \cdot 3(3x - 2)² \cdot 3 + (3x - 2)³(4x + 5)⁴ \cdot 2(2x + 1) \cdot 2

= (2x + 1)(3x - 2)²(4x + 5)³[16(2x + 1)(3x - 2) + 9(2x + 1)(4x + 5) + 4(3x - 2)(4x + 5)]

= (2x + 1)(3x - 2)²(4x + 5)³[16(6x² - x - 2) + 9(8x² + 14x + 5) + 4(12x² + 7x - 10)]

= (2x + 1)(3x - 2)²(4x + 5)³(216x² + 138x - 27)

Example 2. Differentiate the following functions:
(i) \( \frac{3x - 2}{5x^2 + 7} \)
(ii) \( \frac{3x + 2}{(x + 5)(2x + 1) + 3} \)
(iii) \( \frac{(1 - 2x)^{5/2}}{2x^2 + 1} \)

Solution. (i) Let y = \( \frac{3x - 2}{5x^2 + 7} \). Differentiating with respect to x, we get
\( \frac{dy}{dx} = \frac{(5x^2 + 7) \cdot \frac{d}{dx}(3x - 2) - (3x - 2) \cdot \frac{d}{dx}(5x^2 + 7)}{(5x^2 + 7)^2} \) (using quotient rule)

= \( \frac{(5x^2 + 7)(3) - (3x - 2)(10x)}{(5x^2 + 7)^2} \)

= \( \frac{15x^2 + 21 - 30x^2 + 20x}{(5x^2 + 7)^2} = -\frac{15x^2 - 20x - 21}{(5x^2 + 7)^2} \)

(ii) Let y = \( \frac{3x + 2}{(x + 5)(2x + 1) + 3} = \frac{3x + 2}{2x^2 + 11x + 8} \). Differentiating with respect to x,
\( \frac{dy}{dx} = \frac{(2x^2 + 11x + 8)(3) - (3x + 2)(4x + 11)}{(2x^2 + 11x + 8)^2} \)

= \( \frac{3(2x^2 + 11x + 8) - (3x + 2)(4x + 11)}{(2x^2 + 11x + 8)^2} \)

= \( \frac{6x^2 + 33x + 24 - (12x^2 + 41x + 22)}{(2x^2 + 11x + 8)^2} = \frac{-6x^2 - 8x + 2}{(2x^2 + 11x + 8)^2} \)

= \( -\frac{2(3x^2 + 4x - 1)}{(2x^2 + 11x + 8)^2} \)

(iii) Let y = \( \frac{(1 - 2x)^{5/2}}{2x^2 + 1} \). Differentiating with respect to x, we get
\( \frac{dy}{dx} = \frac{(2x^2 + 1) \cdot \frac{5}{2}(1 - 2x)^{3/2} \cdot (-2) - (1 - 2x)^{5/2} \cdot (4x)}{(2x^2 + 1)^2} \)

= \( \frac{(1 - 2x)^{3/2}[-5(2x^2 + 1) - 4x(1 - 2x)]}{(2x^2 + 1)^2} \)

= \( \frac{(1 - 2x)^{3/2}(-10x^2 - 5 - 4x + 8x^2)}{(2x^2 + 1)^2} = \frac{(1 - 2x)^{3/2}(-2x^2 - 4x - 5)}{(2x^2 + 1)^2} \)

= \( -\frac{(1 - 2x)^{3/2}(2x^2 + 4x + 5)}{(2x^2 + 1)^2} \)

Example 3. Differentiate \( \frac{4x^2 - 1}{(5 - 2x)^3} \) and find the value of the derivative at x = 2.
Solution: Let f(x) = \( \frac{4x^2 - 1}{(5 - 2x)^3} \). Differentiating with respect to x, we get
\( f'(x) = \frac{(5 - 2x)^3 \cdot (8x) - (4x^2 - 1) \cdot 3(5 - 2x)^2 \cdot (-2)}{((5 - 2x)^3)^2} \)

= \( \frac{(5 - 2x)^2[(5 - 2x) \cdot 8x + 6(4x^2 - 1)]}{(5 - 2x)^6} = \frac{8x^2 + 40x - 6}{(5 - 2x)^4} \)

\( \therefore f'(2) = \frac{8(4) + 40(2) - 6}{(5 - 4)^4} = \frac{32 + 80 - 6}{1} = 106 \)

Example 4. If y = \( \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} \), prove that \( 2xy \frac{dy}{dx} = \frac{x}{a} - \frac{a}{x} \).
Solution: Given y = \( \frac{1}{\sqrt{a}} \cdot x^{1/2} + \sqrt{a} \cdot x^{-1/2} \). Differentiating with respect to x, we get
\( \frac{dy}{dx} = \frac{1}{\sqrt{a}} \cdot \frac{1}{2} x^{-1/2} + \sqrt{a} \cdot \left(-\frac{1}{2}\right) x^{-3/2} = \frac{1}{2\sqrt{ax}} - \frac{\sqrt{a}}{2x^{3/2}} \)

\( \implies 2x \frac{dy}{dx} = \frac{\sqrt{x}}{\sqrt{a}} - \frac{\sqrt{a}}{\sqrt{x}} \)

\( \implies 2xy \frac{dy}{dx} = \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right) \left(\frac{\sqrt{x}}{\sqrt{a}} - \frac{\sqrt{a}}{\sqrt{x}}\right) = \frac{x}{a} - \frac{a}{x} \)

Example 5. Find the coordinates of the points on the curve y = \( \frac{x}{1 - x^2} \) for which \( \frac{dy}{dx} = 1 \).
Solution: Given y = \( \frac{x}{1 - x^2} \) \( \implies \)
\( \frac{dy}{dx} = \frac{(1 - x^2) \cdot 1 - x \cdot (-2x)}{(1 - x^2)^2} = \frac{1 + x^2}{(1 - x^2)^2} \)

Now \( \frac{dy}{dx} = 1 \) \( \implies \) \( \frac{1 + x^2}{(1 - x^2)^2} = 1 \)

\( \implies x^4 - 3x^2 = 0 \) \( \implies x^2(x^2 - 3) = 0 \) \( \implies x = 0, \pm\sqrt{3} \)

When x = 0, y = 0;

When x = √3, y = \( \frac{\sqrt{3}}{1 - 3} = -\frac{\sqrt{3}}{2} \); and

When x = -√3, y = \( \frac{-\sqrt{3}}{1 - 3} = \frac{\sqrt{3}}{2} \)

Hence, the points are (0, 0), \( \left(\sqrt{3}, -\frac{\sqrt{3}}{2}\right) \), \( \left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right) \)

Exercise 6.1

Differentiate the following (1 to 6) functions:

Question 1. (i) (2x + 3)(5x² - 7x + 1)
(ii) (3x⁴ - 5)(7x³ - 11x + 2)

Question 2. (i) x³√(3x - 4)
(ii) x(x - 2)√(x - 3)

Question 3. (i) (x + 1)(5x + 7)²(2x + 3)³
(ii) x²(3x + 2)³(1 - 2x)⁴

Question 4. (i) \( \frac{x}{3x^2 + 5} \)
(ii) \( \frac{(2x + 1)(3x - 1)}{x + 5} \)

Question 5. (i) \( \frac{ax^2 + bx + c}{px^2 + qx + r} \)
(ii) \( \frac{1}{\sqrt{2ax + bx + c}} \) [Note: Likely should be \( \frac{1}{\sqrt{ax^2 + bx + c}} \)]

Question 6. (i) \( \frac{(2 + 5x)^2}{x^3 - 1} \)
(ii) \( \sqrt{\frac{1 + x}{1 - x}} \)

Question 7. Differentiate \( \frac{2x^2 - 4}{3x^2 + 7} \) and find the value of the derivative at x = 1.

Question 8. If y = √x + \( \frac{1}{\sqrt{x}} \), prove that \( 2x \frac{dy}{dx} + y = 2\sqrt{x} \).

Question 9. If y = \( \frac{1}{\sqrt{x}}\left(1 - \frac{1}{x}\right) \), prove that \( x^{5/2}\left(2\frac{dy}{dx} - y\right) + x^2 - 3 = 0 \).

Question 10. If y = \( \frac{x}{x + a} \), prove that \( x\frac{dy}{dx} = y(1 - y) \).

Question 11. Given y = (3x - 1)² + (2x - 1)³, find \( \frac{dy}{dx} \) and the points on the curve for which \( \frac{dy}{dx} = 0 \).

6.2 Derivatives of Trigonometric Functions

1. Derivative of sin x

Let f(x) = sin x, then f(x + δx) = sin(x + δx).

By definition, f'(x) = \( \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} = \lim_{\delta x \to 0} \frac{\sin(x + \delta x) - \sin x}{\delta x} \)

= \( \lim_{\delta x \to 0} \frac{2\cos\left(x + \frac{\delta x}{2}\right) \sin\left(\frac{\delta x}{2}\right)}{\delta x} \) (C - D formulae)

= \( \lim_{\delta x \to 0} \frac{2\cos\left(x + \frac{\delta x}{2}\right) \sin\left(\frac{\delta x}{2}\right)}{\delta x} = \lim_{\delta x/2 \to 0} \cos\left(x + \frac{\delta x}{2}\right) \cdot \lim_{\delta x/2 \to 0} \frac{\sin\frac{\delta x}{2}}{\frac{\delta x}{2}} \)

= cos x \cdot 1 = cos x.

Thus, \( \frac{d}{dx}(\sin x) = \cos x \), for all x \( \in \) R.

2. Derivative of cos x

Let f(x) = cos x, then f(x + δx) = cos(x + δx).

By definition, f'(x) = \( \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} = \lim_{\delta x \to 0} \frac{\cos(x + \delta x) - \cos x}{\delta x} \)

= \( \lim_{\delta x \to 0} \frac{2\sin\left(x + \frac{\delta x}{2}\right) \sin\left(\frac{-\delta x}{2}\right)}{\delta x} \) (C - D formulae)

= \( \lim_{\delta x \to 0} \frac{2\sin\left(x + \frac{\delta x}{2}\right) \sin\left(-\frac{\delta x}{2}\right)}{\delta x} \)

= \( -\lim_{\delta x/2 \to 0} \sin\left(x + \frac{\delta x}{2}\right) \cdot \lim_{\delta x/2 \to 0} \frac{\sin\frac{\delta x}{2}}{\frac{\delta x}{2}} \)

= - sin x \cdot 1 = - sin x.

Thus, \( \frac{d}{dx}(\cos x) = -\sin x \), for all x \( \in \) R.