Question

Assertion : If $u=f(\tan x),v=g(\sec x)$ and $f^{\prime }(1)=2$
$g(\sqrt{2})=4$, then ${\left(\frac{du}{dv}\right)}_{x=\pi /4}=\frac{1}{\sqrt{2}}$
Reason : If $u=f(x),v=g(x)$, then the derivative of $f$ with respect to $g$ is $\frac{du}{dv}=\frac{du/dx}{dv/dx}$

• A. Assertion is correct, reason is correct; reason is a correct explanation for assertion.
• B. Assertion is correct, reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, reason is incorrect
• D. Assertion is incorrect, reason is correct.

Answer 1

Answer: (A)

Given, $u=f(\tan x)$
$\Rightarrow \frac{du}{dx}=f^{\prime }(\tan x){\sec }^{2}x$
and $v=g(\sec x)$
$\Rightarrow \frac{dv}{dx}=g^{\prime }(\sec x)\sec x\tan x$
$\therefore \frac{du}{dv}=\frac{(du/dx)}{(dv/dx)}=\frac{f^{\prime }(\tan x)}{g^{\prime }(\sec x)}\cdot \frac{1}{\sin x}$
$\therefore {\left(\frac{du}{dv}\right)}_{x=\pi /4}=\frac{f^{\prime }(1)}{g^{\prime }(\sqrt{2})}\cdot \sqrt{2}=\frac{1}{2}\sqrt{2}=\frac{1}{\sqrt{2}}$