Question

The derivative of ${\sec }^{-1}\left(\frac{1}{2x^2+1}\right)$ with respect to $\sqrt{1+3x}$ at $x=-1/3$

• A. $doesnotexist$
• B. $0$
• C. $1/2$
• D. $1/3$

Answer 1

The correct option is B $0$

We have,

${\sec }^{-1}\left(\frac{1}{2x^2+1}\right)$ with respect to $\sqrt{1+3x}$

Now, Let $u={\sec }^{-1}\left(\frac{1}{2x^2+1}\right)andv=\sqrt{1+3x}$

Put $x=\cos \theta ,\theta ={\cos }^{-1}x$

So,

$u={\sec }^{-1}\left(\frac{1}{2{\cos }^2\theta -1}\right)$

$u={\sec }^{-1}\left(\frac{1}{\cos 2\theta }\right)$

$u={\sec }^{-1}\sec 2\theta$

$u=2\theta$

$u=2{\cos }^{-1}x$

On differentiating with respect to x and we get,

$\frac{du}{dx}=-2\frac{1}{\sqrt{1-x^2}}$

Now,

$v=\sqrt{1+3x}$

On differentiating this with respect t x and we get,

$\frac{dv}{dx}=\frac{1}{2\sqrt{1+3x}}\frac{d}{dx}(1+3x)$

$\frac{dv}{dx}=\frac{3}{2\sqrt{1+3x}}$

Now,

According to given question,

$\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{-2}{\sqrt{1-x^2}}}{\frac{3}{2\sqrt{1+3x}}}=\frac{-4\sqrt{1+3x}}{3\sqrt{1-x^2}}$

At the point $x=\frac{-1}{3}$

Then,

$\frac{du}{dv}=\frac{-4\sqrt{1+3\times \left(\frac{-1}{3}\right)}}{3\sqrt{1-{\left(\frac{-1}{3}\right)}^2}}$

$\frac{du}{dv}=\frac{-4\sqrt{1-1}}{3\sqrt{\frac{9-1}{9}}}$

$\frac{du}{dv}=0$

Hence, this is the answer.