Question

Find derivative of the following function:

\ln [\tan (\frac{π}{4} + \frac{x}{2})]

.

Accepted Answer

Let $y = f (x) = \ln [\tan (\frac{π}{4} + \frac{x}{2})]$

Now, differentiate $y$ with respect to $x$ , then we get: $f^{'} (x) = \frac{d y}{d x} = \frac{d}{d x} \{\ln [\tan (\frac{π}{4} + \frac{x}{2})]\}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\tan (\frac{π}{4} + \frac{x}{2})} \frac{d}{d x} [\tan (\frac{π}{4} + \frac{x}{2})]$ [Using chain rule of differentiation]

$\Rightarrow \frac{d y}{d x} = \frac{1}{\tan (\frac{π}{4} + \frac{x}{2})} \times \sec^{2} (\frac{π}{4} + \frac{x}{2}) \frac{d}{d x} (\frac{π}{4} + \frac{x}{2})$ [Using $\frac{d}{d x} (\tan (a x + b)) = a \cdot \sec^{2} (a x + b)$ ]

$\Rightarrow \frac{d y}{d x} = \frac{1}{\tan (\frac{π}{4} + \frac{x}{2})} \times \sec^{2} (\frac{π}{4} + \frac{x}{2}) \times \frac{1}{2}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\cos^{2} (\frac{π}{4} + \frac{x}{2})} \times \frac{\cos (\frac{π}{4} + \frac{x}{2})}{\sin (\frac{π}{4} + \frac{x}{2})} \times \frac{1}{2}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 \cos (\frac{π}{4} + \frac{x}{2}) \sin (\frac{π}{4} + \frac{x}{2})}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\sin (2 (\frac{π}{4} + \frac{x}{2}))}$ [Using $\sin 2 A = 2 \sin A \cos A$ ]

$\Rightarrow \frac{d y}{d x} = \frac{1}{\sin (\frac{π}{2} + x)}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\cos x}$

$\Rightarrow \frac{d y}{d x} = \sec x$

Hence, the derivative of $\ln [\tan (\frac{π}{4} + \frac{x}{2})]$ is $\sec x$ .

Odisha Board Solutions for Chapter: Continuity and Differentiability, Exercise 3: EXERCISE 7(c)

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Continuity and Differentiability, Exercise 3: EXERCISE 7(c)

Questions from Odisha Board Solutions for Chapter: Continuity and Differentiability, Exercise 3: EXERCISE 7(c) with Hints & Solutions

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