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

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(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

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