First slide

Derivatives

Question

$If y = {Tan}^{- 1} (\sec x + \tan x) then \frac{d y}{d x} =$

Moderate

A

1
B

$\frac{1}{2}$
C

-1
D

0

Solution

$\begin{aligned} y = \tan^{- 1} (\sec x + \tan x) \\ y = \tan^{- 1} (\frac{1}{\cos x} + \frac{\sin x}{\cos x}) \\ y = \tan^{- 1} \frac{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}}{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}} \\ y = \tan^{- 1} (\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}) \end{aligned}$
$\begin{array}{l} y = \tan^{- 1} (\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}) \\ y = \tan^{- 1} (\tan (\frac{π}{4} + \frac{x}{2})) \\ y = \frac{π}{4} + \frac{x}{2} \\ \frac{d y}{d x} = 0 + \frac{1}{2} \\ \frac{d y}{d x} = \frac{1}{2} \end{array}$

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