First slide

Derivatives

Question

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

Moderate

A

1
B

$\sqrt{2}$
C

$\frac{1}{\sqrt{2}}$
D

$\frac{1}{2}$

Solution

$y = \sec (T a n^{- 1} x) = \sqrt{\sec^{2} \tan^{- 1} x} = \sqrt{1 + \tan^{2} \tan^{- 1} x} = \sqrt{1 + x^{2}}$
$\frac{d y}{d x} = \frac{2 x}{2 \sqrt{1 + x^{2}}} = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \frac{1}{\sqrt{2}}$

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