If u=$f\left(x^3\right)$,  v=$g\left(x^2\right),$  $f^{\text{'}}\left(x\right)$=$\cos x$ and $g\text{'}\left(x\right)$=$\sin x \mathrm{then} \frac{d\mathrm{u}}{d\mathrm{v}} \mathrm{is}:$ 

$$\begin{gathered} \mathrm{u}=\mathrm{f}\left({\mathrm{x}}^3\right), \mathrm{v}=\mathrm{g}\left({\mathrm{x}}^2\right) {\mathrm{f}}^1\left(\mathrm{x}\right)=\cos \mathrm{x} \\ {\mathrm{g}}^1\left(\mathrm{x}\right)=\sin \mathrm{x} \\ \frac{\mathrm{du}}{\mathrm{dv}}=\frac{{\mathrm{f}}^1\left({\mathrm{x}}^3\right)3{\mathrm{x}}^2}{{\mathrm{g}}^1\left({\mathrm{x}}^2\right) 2\mathrm{x}} \\ =\frac{3x \cos {\mathrm{x}}^3}{2 \sin {\mathrm{x}}^2} \\ =\frac{3}{2}x \cos {\mathrm{x}}^3 \mathrm{cosec} {\mathrm{x}}^2 \end{gathered}$$