Question

If $f^{\prime }\left(x\right)=\ln x$ and $y=f\left(\sec \sqrt{x}\right),$ then derivative of $y$ with respect to ${\sin }^2\sqrt{x}$ is

• A. $\ln \left(\sec \sqrt{x}\right)\cdot {\sec }^3\sqrt{x}$
• B. $2\ln \left(\sec \sqrt{x}\right)\cdot {\sec }^2\sqrt{x}\cdot \sin \sqrt{x}$
• C. $\ln \left(\sec \sqrt{x}\right)\cdot {\sec }^2\sqrt{x}$
• D. $\frac{1}{2}\ln \left(\sec \sqrt{x}\right)\cdot {\sec }^3\sqrt{x}$

Answer 1

The correct option is D $\frac{1}{2}\ln \left(\sec \sqrt{x}\right)\cdot {\sec }^3\sqrt{x}$

Let $g\left(x\right)={\sin }^2\sqrt{x}$
Then we have to find $\frac{dy}{dg}.$
$\frac{dy}{dg}=\frac{dy/dx}{dg/dx}\dots \left(1\right)$

$y=f\left(\sec \sqrt{x}\right)$
$\Rightarrow y^{\prime }\left(x\right)=f^{\prime }\left(\sec \sqrt{x}\right)\frac{d\left(\sec \sqrt{x}\right)}{dx}$
Given that $f^{\prime }\left(x\right)=\ln x$
$y^{\prime }\left(x\right)=\ln \left(\sec \sqrt{x}\right)\cdot \left(\sec \sqrt{x}\right)\cdot \left(\tan \sqrt{x}\right)\cdot \frac{1}{2\sqrt{x}}$
$=\frac{1}{2\sqrt{x}}\cdot \ln \left(\sec \sqrt{x}\right)\cdot \left(\sin \sqrt{x}\right)\cdot \left({\sec }^2\sqrt{x}\right)$

and $g^{\prime }\left(x\right)=2\left(\sin \sqrt{x}\right)\cdot \left(\cos \sqrt{x}\right)\cdot \frac{1}{2\sqrt{x}}$
Now from equation $\left(1\right),$ we have
$\frac{dy}{dg}=\frac{\frac{1}{2\sqrt{x}}\cdot \ln \left(\sec \sqrt{x}\right)\cdot \left(\sin \sqrt{x}\right)\cdot \left({\sec }^2\sqrt{x}\right)}{\left(\sin \sqrt{x}\right)\cdot \left(\cos \sqrt{x}\right)\cdot \frac{1}{\sqrt{x}}}=\frac{\frac{1}{2}\ln \left(\sec \sqrt{x}\right)}{{\cos }^3\sqrt{x}}$