Q81 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
The derivative of $f(\sin x)$ with respect to $g(\sec x)$ at $x = \dfrac{\pi}{4}$, given that $f'\left(\dfrac{1}{\sqrt{2}}\right) = 3$ and $g'(\sqrt{2}) = 1$ is.....
- A$\dfrac{\sqrt{3}}{2}$
- Correct answerB$\dfrac{3}{2}$
- C$\dfrac{1}{3}$
- D$\dfrac{-1}{3}$
Solution
Let $u = f(\sin x)$ and $v = g(\sec x)$
$\dfrac{du}{dx} = f'(\sin x) \cos x$
$\dfrac{dv}{dx} = g'(\sec x) \sec x \tan x$
$\dfrac{du}{dv} = \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}} = \dfrac{f'(\sin x) \cos x}{g'(\sec x) \sec x \tan x}$
At $x = \dfrac{\pi}{4}$:
$\dfrac{du}{dv} = \dfrac{f'\left(\dfrac{1}{\sqrt{2}}\right) \cdot \dfrac{1}{\sqrt{2}}}{g'(\sqrt{2}) \cdot \sqrt{2} \cdot 1}$
Substituting $f'\left(\dfrac{1}{\sqrt{2}}\right) = 3$ and $g'(\sqrt{2}) = 1$:
$\dfrac{du}{dv} = \dfrac{3 \cdot \dfrac{1}{\sqrt{2}}}{1 \cdot \sqrt{2}} = \dfrac{3}{2}$
Answer: $\dfrac{3}{2}$
$\dfrac{du}{dx} = f'(\sin x) \cos x$
$\dfrac{dv}{dx} = g'(\sec x) \sec x \tan x$
$\dfrac{du}{dv} = \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}} = \dfrac{f'(\sin x) \cos x}{g'(\sec x) \sec x \tan x}$
At $x = \dfrac{\pi}{4}$:
$\dfrac{du}{dv} = \dfrac{f'\left(\dfrac{1}{\sqrt{2}}\right) \cdot \dfrac{1}{\sqrt{2}}}{g'(\sqrt{2}) \cdot \sqrt{2} \cdot 1}$
Substituting $f'\left(\dfrac{1}{\sqrt{2}}\right) = 3$ and $g'(\sqrt{2}) = 1$:
$\dfrac{du}{dv} = \dfrac{3 \cdot \dfrac{1}{\sqrt{2}}}{1 \cdot \sqrt{2}} = \dfrac{3}{2}$
Answer: $\dfrac{3}{2}$
Hence, the correct answer is option: 'B'.