Question

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

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

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the value of $\frac{dy}{dx}$ at $x=1$:

Given,

$y=sec\left({\tan }^{-1}x\right)$

We can write ${\tan }^{-1}x\mathrm{as}{\sec }^{-1}\sqrt{1+{\mathrm{x}}^2}$

So,

$$\begin{gathered} y=sec\left(se{c}^{-1}\sqrt{1+x^2}\right) \\ y=\sqrt{1+x^2} \end{gathered}$$

Now differentiate with respect to $x$

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{2\sqrt{1+x^2}}\left(2x\right)\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\sqrt{\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{x}}}\mathbf{]}\\ & =& \frac{x}{\sqrt{1+x^2}}\end{array}$

Now at $x=1$,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{\sqrt{1+1^2}}\\ & =& \frac{1}{\sqrt{2}}\end{array}$

Hence, the correct option is A.