Question

If $y={(xco{t}^{3}x)}^{3/2}$, then $\frac{dy}{dx}=?$

• A. $\frac{3}{2}{\left(x{\cot }^{3}x\right)}^{1/2}\left[{\cot }^{3}x-3x{\cot }^{2}x{\mathrm{cosec}}^{2}x\right]$
• B. $\frac{3}{2}{\left(x{\cot }^{3}x\right)}^{1/2}\left[{\cot }^{2}x-3x{\cot }^{2}x{\mathrm{cosec}}^{2}x\right]$
• C. $\frac{3}{2}{\left(x{\cot }^{3}x\right)}^{3/2}\left[{\cot }^{3}x-3x{\mathrm{cosec}}^{2}x\right]$
• D. None of these

Answer 1

The correct option is A

$\frac{3}{2}{\left(x{\cot }^{3}x\right)}^{1/2}\left[{\cot }^{3}x-3x{\cot }^{2}x{\mathrm{cosec}}^{2}x\right]$

Explanation for the correct option:

Find the first derivative:

We have been given $y={(xco{t}^{3}x)}^{3/2}$.

Differentiating with respect to $x$ we get,

$\begin{array}{rcl}\frac{dy}{dx}& =& {\frac{3}{2}(xco{t}^{3}x)}^{1/2}\left[x\frac{d}{dx}\left(co{t}^{3}x\right)+\left(co{t}^{3}x\right)\frac{d}{dx}\left(x\right)\right]\\ & =& {\frac{3}{2}(xco{t}^{3}x)}^{1/2}\left[x\left\{\left(3co{t}^{2}x\right)·\left(-\cos e{c}^{2}x\right)\right\}+\left(co{t}^{3}x\right)\right]\\ & =& {\frac{3}{2}(xco{t}^{3}x)}^{1/2}\left[co{t}^{3}x-3xco{t}^{2}x\cos e{c}^{2}x\right]\end{array}$

Hence, option (A) is correct.