Question

differentiate log (1+x2) with respect to cot-1 x​

Answer 1

$\huge \bf{ \red{ \underline{Answer}}} \\ = \bf{- 2x }\\ \\ \bf{ \underline { step \: by \: step \: solution}} \\ \\ let \\ \bf{y = log \: (1 + {x}^{2} )} \\ \\ \bf{and \: } \\ \bf{z = {cot}^{ - 1} x} \\ \\ \\ \red{ \bf{find \:} } - \bf{differentiate \: to \: y\: with \: respect \: } \\ \\ \bf{ to \: z} \\ \huge \: \bf{ solution \: } \\ \\ \bf{we \: have \: } \\ \bf{y = log(1 + {x}^{2} )} \\ \bf{differentiating \: with \: respect \: to \: x \: } \\ \\ \bf{\frac{dy}{dx} = \frac{1}{1 + {x}^{2} } \times 2x} \\ \\ \bf{now \: we \: have \: } \\ \bf{z }= {cot}^{ - 1} x \\ \bf{differentiating \: with \: respect \: to \: x \: } \\ \\ \bf{\frac{dz}{ dx} = \frac{ - 1}{1 + {x}^{2} } } \\ \\ \bf{now }\: \\ \bf{we \: have \: to \: find} \: \\ \\ \bf{\frac{dy}{dz} = \frac{ \frac{dy}{dx} }{ \frac{dz}{dx} } } \\ \\ \bf{ \frac{dy}{dz} = \frac{ 2x}{1 + {x}^{2} } \times \frac{1 + {x}^{2} }{ - 1}} \\ \\ \bf{\frac{dy}{dz} = - 2x} \\ \\ \bf{this \: is \: the \: required \: solution}$