If $u = f (x^{3}), v = g (x^{2}), f' (x) = cosx$ and $g' (x) = sinx$ then $\frac{du}{dv}$ is

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$\frac{3}{2} x . {cosx}^{3} . {cosecx}^{2}$

sec x

$\frac{2}{3} \sin x^{3} . {secx}^{3}$

tan x

answer is A.

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Detailed Solution

Given $u = f (x^{3})$ and $V = g (x^{2})$

Given $f' (x) = \cos x, g' (x) = \sin x$

$\begin{array}{rcl} \frac{d u}{d v} & = & \frac{f' (x^{3}) 3 x^{2}}{g' (x^{2}) 2 x} \\ = & \frac{\cos (x^{3}) 3 x}{2 \sin (x^{2})} \\ = & \frac{3}{2} x \cos (x^{3}) \cos e c (x^{2}) \end{array}$

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