Question

Assertion : If $u=f(\tan x), v=g(\sec x)$ and $f'(1)=2$
$g(\sqrt{2})=4$, 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) $
$\Rightarrow \frac{d u}{d x}=f^{\prime}(\tan x) \sec ^{2} x$
and $v=g(\sec x)$
$\Rightarrow \frac{d v}{d x}=g'(\sec x) \sec x \tan x$
$\therefore \frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{f'(\tan x)}{g^{\prime}(\sec x)} \cdot \frac{1}{\sin x}$
$\therefore \left(\frac{d u}{d v}\right)_{x=\pi / 4}=\frac{f'(1)}{g'(\sqrt{2})} \cdot \sqrt{2}=\frac{1}{2} \sqrt{2}=\frac{1}{\sqrt{2}}$