Question

If y = $sec\left(ta{n}^{-1}x\right)$, then $\frac{dy}{dx}$ at x = 1 is equal to:

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $1$
• D. $0$

Answer 1

The correct option is A

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

Let $y=sec\left(ta{n}^{-1}x\right)andta{n}^{-1}x=\theta .\Rightarrow x=tan\theta \Rightarrow y=\sqrt{1+x^2}(\because se{c}^2\theta =1+ta{n}^2\theta )\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{1+x^2}}.2xAtx=1,\frac{dy}{dx}=\frac{1}{\sqrt{2}}.$