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 $y = f(	an x)$ and $z = g(\sec x)$. We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.First,find the derivative of $y$ with respect to $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $y = f(	an x)$ and $z = g(\sec x)$. We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.First,find the derivative of $y$ with respect to $x$:$\frac{dy}{dx} = f^{\prime}(	an x) \cdot \sec^2 x$.Next,find the derivative of $z$ with respect to $x$:$\frac{dz}{dx} = g^{\prime}(\sec x) \cdot \sec x 	an x$.Now,calculate the ratio at $x = \frac{\pi}{4}$:$\frac{dy}{dz} = \frac{f^{\prime}(	an x) \cdot \sec^2 x}{g^{\prime}(\sec x) \cdot \sec x 	an x} = \frac{f^{\prime}(	an x) \cdot \sec x}{g^{\prime}(\sec x) \cdot 	an x}$.At $x = \frac{\pi}{4}$,$	an x = 1$ and $\sec x = \sqrt{2}$:$\frac{dy}{dz} = \frac{f^{\prime}(1) \cdot \sqrt{2}}{g^{\prime}(\sqrt{2}) \cdot 1} = \frac{2 \cdot \sqrt{2}}{4 \cdot 1} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

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

Different conformations of the same molecule are called

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch $S$ and a resistor $R$ as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be

Which type of chromosome is shown in the figure below?

If ${a^{x - 1}} = bc$,${b^{y - 1}} = ca$,and ${c^{z - 1}} = ab$,then find the value of $\sum \frac{1}{x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module