The derivative of cot ^{-1} x with respect to | vedclass

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

Question

(B) Let $u = \cot ^{-1} x$ and $v = \log (1+x^{2})$.First,we find the derivative of $u$ with respect to $x$:$\frac{du}{dx} = -\frac{1}{1+x^{2}}$.Next,

Vedclass · Accepted Answer

$-\frac{1}{2 x}$

Vedclass · Answer

(B) Let $u = \cot ^{-1} x$ and $v = \log (1+x^{2})$.First,we find the derivative of $u$ with respect to $x$:$\frac{du}{dx} = -\frac{1}{1+x^{2}}$.Next,we find the derivative of $v$ with respect to $x$ using the chain rule:$\frac{dv}{dx} = \frac{1}{1+x^{2}} 	imes \frac{d}{dx}(1+x^{2}) = \frac{2x}{1+x^{2}}$.Now,we find the derivative of $u$ with respect to $v$:$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-1/(1+x^{2})}{2x/(1+x^{2})} = -\frac{1}{2x}$.Thus,the correct option is $B$.

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

Similar Questions

If $x=\cos \theta$ and $y=\sin 5 \theta$,then $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)

If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$,then $\frac{dy}{dx}$ is equal to

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

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

If $x = a \cos^4 \theta$ and $y = a \sin^4 \theta$,then $\frac{dy}{dx}$ at $\theta = \frac{3\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module