Find the derivative of f(sin x) with respect | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $u = f(\sin x)$ and $v = g(\sec x)$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.Using the chain rule, $\frac{du}{dx} = f'(\sin x) \

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) Let $u = f(\sin x)$ and $v = g(\sec x)$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.Using the chain rule, $\frac{du}{dx} = f'(\sin x) \cdot \cos x$ and $\frac{dv}{dx} = g'(\sec x) \cdot \sec x 	an x$.At $x = \frac{\pi}{4}$, $\sin x = \frac{1}{\sqrt{2}}$, $\cos x = \frac{1}{\sqrt{2}}$, $\sec x = \sqrt{2}$, and $	an x = 1$.Substituting these values: $\frac{du}{dx} = f'(\frac{1}{\sqrt{2}}) \cdot \frac{1}{\sqrt{2}} = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.$\frac{dv}{dx} = g'(\sqrt{2}) \cdot \sqrt{2} \cdot 1 = 1 \cdot \sqrt{2} = \sqrt{2}$.Therefore, $\frac{du}{dv} = \frac{3/\sqrt{2}}{\sqrt{2}} = \frac{3}{2}$.

Find the derivative of $f(\sin x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$, given that $f'(\frac{1}{\sqrt{2}}) = 3$ and $g'(\sqrt{2}) = 1$.

Similar Questions

If the equation of the normal to the curve $x=\sqrt{t}$,$y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

The derivative of $\cot ^{-1} x$ with respect to $\log (1+x^{2})$ is

Find $\frac{dy}{dx}$,if $x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.

What is the slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$?

If $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$ for $|t| \ge 1$,then $\frac{dy}{dx}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module