Question

Differentiate

\frac{\sec x - 1}{\sec x + 1}

with respect to

x

.

Accepted Answer

The given function is $\frac{\sec x - 1}{\sec x + 1}$

Let $y = \frac{\sec x - 1}{\sec x + 1}$

$y = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1} [∵ \frac{1}{\cos x} = \sec x]$

$y = \frac{1 - \cos x}{1 + \cos x}$

On differentiating both sides with respect to $x$ , we get

$\frac{d}{d x} (y) = \frac{d}{d x} (\frac{1 - \cos x}{1 + \cos x})$

$\frac{d y}{d x} = \frac{(1 + \cos x) \frac{d}{d x} (1 - \cos x) - (1 - \cos x) \frac{d}{d x} (1 + \cos x)}{{(1 + \cos x)}^{2}}$ [Using quotient rule]

$\frac{d y}{d x} = \frac{(1 + \cos x) [\frac{d}{d x} (1) - \frac{d}{d x} (\cos x)] - (1 - \cos x) (\frac{d}{d x} (1) + \frac{d}{d x} (\cos x))}{{(1 + \cos x)}^{2}}$

$\frac{d y}{d x} = \frac{(1 + \cos x) [0 - (- \sin x)] - (1 - \cos x) (0 + (- \sin x))}{{(1 + \cos x)}^{2}} [∵ \frac{d}{d x} \cos x = - \sin x; \frac{d}{d x} C = 0]$

$\frac{d y}{d x} = \frac{(1 + \cos x) [0 + \sin x] - (1 - \cos x) (0 - \sin x)}{{(1 + \cos x)}^{2}}$

$\frac{d y}{d x} = \frac{(1 + \cos x) (\sin x) - (1 - \cos x) (- \sin x)}{{(1 + \cos x)}^{2}}$

$\frac{d y}{d x} = \frac{(\sin x + \sin x \cos x) - (- \sin x + \sin x \cos x)}{{(1 + \cos x)}^{2}}$

$\frac{d y}{d x} = \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{{(1 + \cos x)}^{2}}$

$\frac{d y}{d x} = \frac{2 \sin x}{{(1 + \cos x)}^{2}}$

M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 5: EXERCISE 5.5

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 5: EXERCISE 5.5

Questions from M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 5: EXERCISE 5.5 with Hints & Solutions

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