The derivative of f(tan x) with respect to g( | vedclass

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

Question

(A) Let $p = f(	an x)$ and $q = g(\sec x)$.Using the chain rule,we find the derivatives with respect to $x$:$\frac{dp}{dx} = f^{\prime}(	an x) \cdot

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) Let $p = f(	an x)$ and $q = g(\sec x)$.Using the chain rule,we find the derivatives with respect to $x$:$\frac{dp}{dx} = f^{\prime}(	an x) \cdot \sec^2 x$$\frac{dq}{dx} = g^{\prime}(\sec x) \cdot \sec x 	an x$At $x = \frac{\pi}{4}$,we have $	an x = 1$ and $\sec x = \sqrt{2}$.$\left. \frac{dp}{dx} ight|_{x=\frac{\pi}{4}} = f^{\prime}(1) \cdot (\sqrt{2})^2 = 2 \cdot 2 = 4$.$\left. \frac{dq}{dx} ight|_{x=\frac{\pi}{4}} = g^{\prime}(\sqrt{2}) \cdot (\sqrt{2} \cdot 1) = 4 \cdot \sqrt{2} = 4\sqrt{2}$.The required derivative is $\frac{dp}{dq} = \frac{dp/dx}{dq/dx} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.

The derivative of $f(\tan x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$,where $f^{\prime}(1) = 2$ and $g^{\prime}(\sqrt{2}) = 4$,is:

Similar Questions

For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals

If $x=f(\theta)$ and $y=g(\theta)$,then $\frac{d^2 y}{d x^2}=$

If $x=3 \tan t$ and $y=3 \sec t$,then the value of $\frac{d^2 y}{d x^2}$ at $t=\frac{\pi}{4}$ is

Find $\frac{dy}{dx}$,if $x = at^2$ and $y = 2at$.

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module