If y = sec (tan–1x), then \frac{dy}{dx} at x = 1 is equal to :

Engineering

Mathematics

Basic Rule of Differentiation

Question

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

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

$\frac{1}{2}$

$\sqrt{2}$

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y = sec (tan^–1x)

$y = \sec (\sec^{- 1} \sqrt{1 + x^{2}})$

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

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