Question

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

• A. $\frac{x}{\sqrt{1+x^{2}}}$
• B. $x \sqrt{1+x^{2}}$
• C. $\sqrt{1+x^{2}}$
• D. $\frac{1}{\sqrt{1+x^{2}}}$
• E. $\frac{x}{1+x^{2}}$

Answer 1

Answer: (A)

$ y =\sec \left(\tan ^{-1} x\right) $
$ =\sec \left(\sec ^{-1} \sqrt{1+x^{2}}\right) $
$ =\sqrt{1+x^{2}} $
On differentiating both sides w.r.t. $X$, we get
$\frac{d y}{d x}=\frac{1}{2 \sqrt{1+x^{2}}}(2 x)$
$\Rightarrow \frac{d y}{d x}=\frac{x}{\sqrt{1+x^{2}}}$