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Q)

Differentiate the following w.r.t x:$\tan^{-1}(\sec x+\tan x),\large \frac{-\pi}{2}<$$x<\large\frac{\pi}{2}$.

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A)
Toolbox:
  • $(1+\sin x)=\sin^2\large\frac{x}{2}+$$\cos^2\large\frac{x}{2}$$+2\sin\large\frac{x}{2}$$\cos\large\frac{x}{2}=$$(\sin\large\frac{x}{2}+$$\cos\large\frac{x}{2})^2$
  • $\cos x=\cos^2\large\frac{x}{2}-$$\sin^2\large\frac{x}{2}$
  • $\large\frac{1+\tan x}{1-\tan x}$$=\tan(\large\frac{\pi}{4}+\frac{x}{2})$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Given : $\tan^{-1}(\sec x+\tan x),\large \frac{-\pi}{2}<$$x<\large\frac{\pi}{2}$.
Consider $\sec x+\tan x$,this can be written as
$\sec x+\tan x=\large\frac{1}{\cos x}+\frac{\sin x}{\cos x}$
$\qquad\qquad\;\;\;\;=\large\frac{1+\sin x}{\cos x}$
Now $(1+\sin x)=\sin^2\large\frac{x}{2}+$$\cos^2\large\frac{x}{2}$$+2\sin\large\frac{x}{2}$$\cos\large\frac{x}{2}=$$(\sin\large\frac{x}{2}+$$\cos\large\frac{x}{2})^2$
$\qquad\qquad\;\;\;\;\;\;\;=(\cos\large\frac{x}{2}+$$\sin\large\frac{x}{2})^2$
$\cos x=\cos^2\large\frac{x}{2}-$$\sin^2\large\frac{x}{2}$
$\quad\quad\;\;=\large\frac{(\cos\Large\frac{x}{2}+\sin\Large\frac{x}{2})^2}{\cos^2\Large\frac{x}{2}-\sin^2\Large\frac{x}{2}}$
$\quad\quad\;\;=\large\frac{(\cos\Large\frac{x}{2}+\sin\Large\frac{x}{2})(\cos\Large\frac{x}{2}+\sin\Large\frac{x}{2})}{(\cos \Large\frac{x}{2}-\sin \Large\frac{x}{2})(\cos \Large\frac{x}{2}+\sin \Large\frac{x}{2})}$
$\quad\quad\;\;=\large\frac{(\cos\Large\frac{x}{2}+\sin\Large\frac{x}{2})}{\cos\Large\frac{x}{2}-\sin\Large\frac{x}{2}}$
Step 2:
Divide throughout by $\cos\large\frac{x}{2}$
$\quad\quad\;\;=\large\frac{1+\tan\Large\frac{x}{2}}{1+\tan\Large\frac{x}{2}}$
But $1=\tan\large\frac{\pi}{4}$
$\quad\quad\;\;=\large\frac{\tan\Large\frac{\pi}{4}+\tan\Large\frac{x}{2}}{1-\tan\Large\frac{\pi}{4}.\tan\Large\frac{x}{2}}$
This is of the form $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\Rightarrow \tan\big(\large\frac{\pi}{4}+\large\frac{x}{2})$
Now $\tan^{-1}(\sec x+\tan x)=\tan^{-1}\big(\tan(\large\frac{\pi}{4}+\frac{x}{2})\big)$
$\Rightarrow \large\frac{\pi}{4}+\frac{x}{2}$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\frac{1}{2}$
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