# How do you differentiate f(x)=cot5x * cot3x using the product rule?

Jul 24, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = - 3 \cot 5 x {\csc}^{2} 3 x - 5 \cot 3 x {\csc}^{2} 5 x$

#### Explanation:

According to product rule, if $f \left(x\right) = g \left(x\right) \times h \left(x\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = g \left(x\right) \times \frac{\mathrm{dh}}{\mathrm{dx}} + \frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right)$

Here as $g \left(x\right) = \cot 5 x$ and $h \left(x\right) = \cot 3 x$, hence

$\frac{\mathrm{df}}{\mathrm{dx}} = \cot 5 x \times - {\csc}^{2} 3 x \times 3 + \left(- {\csc}^{2} 5 x \times 5 \times \cot 3 x\right)$

= $- 3 \cot 5 x {\csc}^{2} 3 x - 5 \cot 3 x {\csc}^{2} 5 x$