Discussion Forum



1)

If $y=\sec(\tan^{-1}x)$   then $\frac{dy}{dx}$  at x=1 is equal to 


A) $\frac{1}{\sqrt{2}}$

B) $\frac{1}{2}$

C) 1

D) $\sqrt{2}$

Answer:

Option A

Explanation:

Given , y= sec {tan-1   x)

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   Let   $\tan^{-1}x=\theta\Rightarrow x=\tan\theta$

   $\therefore$     $y=\sec\theta=\sqrt{1+x^{2}}$

  On differentiating w.r.t x, we get

        $\frac{dy}{dx}=\frac{1}{2\sqrt{1+x^{2}}}2x$

   At x=1,    $\frac{dy}{dx}=\frac{1}{\sqrt{2}}$