The derivative of frac{1-x^2}{1+x^2} with res | vedclass

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

Question

(B) Let $u = \frac{1-x^2}{1+x^2}$ and $v = \frac{2x}{1+x^2}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.First,find $\frac{du}{dx}$:$\frac{d

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(B) Let $u = \frac{1-x^2}{1+x^2}$ and $v = \frac{2x}{1+x^2}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.First,find $\frac{du}{dx}$:$\frac{du}{dx} = \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} = \frac{-4x}{(1+x^2)^2}$.Next,find $\frac{dv}{dx}$:$\frac{dv}{dx} = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2}$.Now,calculate $\frac{du}{dv}$:$\frac{du}{dv} = \frac{-4x}{(1+x^2)^2} \div \frac{2(1-x^2)}{(1+x^2)^2} = \frac{-4x}{2(1-x^2)} = \frac{-2x}{1-x^2} = \frac{2x}{x^2-1}$.At $x=2$:$\frac{du}{dv} = \frac{2(2)}{2^2-1} = \frac{4}{4-1} = \frac{4}{3}$.

The derivative of $\frac{1-x^2}{1+x^2}$ with respect to $\frac{2x}{1+x^2}$ at $x=2$ is

Similar Questions

If $x = \frac{2t}{1 + t^2}$ and $y = \frac{1 - t^2}{1 + t^2}$,then $\frac{dy}{dx}$ equals

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=2at^{2}$ and $y=at^{4}$.

If $x = at^2$ and $y = 2at$,then find $\frac{dy}{dx}$.

For a positive constant $a$,find $\frac{dy}{dx}$,where $y = a^{t+\frac{1}{t}}$ and $x = \left(t+\frac{1}{t}\right)^{a}$.

If $x = a \sin 2\theta (1 + \cos 2\theta )$ and $y = b \cos 2\theta (1 - \cos 2\theta )$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module