Question

Find the derivative of tan^-1 (sec x -tan x)^2

Answer 1

Dear student,

$$\begin{gathered} Letf={\tan }^{-1}{\left(secx-\tan x\right)}^2 \\ Substituteu={\tan }^{-1}\left(secx-\tan x\right) \\ Differentiatingboththesides \\ \frac{df\left(u\right)}{dx}=\frac{df}{du}.\frac{du}{dx} \\ =\frac{d}{du}\left(u^2\right).\frac{d}{dx}\left[{\tan }^{-1}\left(secx-\tan x\right)\right] \\ =2u.\left[\frac{1}{{\left(secx-\tan x\right)}^2+1}\times \frac{d}{dx}\left(secx-\tan x\right)\right] \\ =2u.\left[\frac{1}{{\left(secx-\tan x\right)}^2+1}\times secx\tan x-se{c}^{2}x\right] \\ =2u.\left[\frac{secx\tan x-se{c}^{2}x}{{\left(secx-\tan x\right)}^2+1}\right] \\ Undosubstituiton \\ =2{\tan }^{-1}\left(secx-\tan x\right).\left[\frac{secx\tan x-se{c}^{2}x}{{\left(secx-\tan x\right)}^2+1}\right] \\ =\frac{2{\tan }^{-1}\left(secx-\tan x\right)\left(secx\tan x-se{c}^{2}x\right)}{{\left(secx-\tan x\right)}^2+1} \end{gathered}$$

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