Question
Diffrentiate the following w. r. t. x.
$\cot ^{-1}\left(x^3\right)$
$\cot ^{-1}\left(x^3\right)$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\cot ^{-1}\left(x^3\right)\right] \\ & =\frac{-1}{1+\left(x^3\right)^2} \cdot \frac{d}{d x}\left(x^3\right) \\ & =\frac{-1}{1+x^6} \times 3 x^2 \\ & =\frac{-3 x^2}{1+x^6}\end{aligned}$
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