# How do you differentiate f(x)=(x^3-x)(cotx-2x) using the product rule?

Mar 21, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = - \left({\csc}^{2} x + 8\right) {x}^{3} + 3 \cot x \cdot {x}^{2} + \left({\csc}^{2} x + 4\right) x - \cot x$

#### Explanation:

Product rule states that if $f \left(x\right) = g \left(x\right) \times h \left(x\right)$, then

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

Here $g \left(x\right) = \left({x}^{3} - x\right)$ and $h \left(x\right) = \left(\cot x - 2 x\right)$

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = \left({x}^{3} - x\right) \times \left(- {\csc}^{2} x - 2\right) + \left(\cot x - 2 x\right) \times \left(3 {x}^{2} - 1\right)$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = - {x}^{3} {\csc}^{2} x + x {\csc}^{2} x - 2 {x}^{3} + 2 x + 3 {x}^{2} \cot x - 6 {x}^{3} - \cot x + 2 x$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = - \left({\csc}^{2} x + 8\right) {x}^{3} + 3 \cot x \cdot {x}^{2} + \left({\csc}^{2} x + 4\right) x - \cot x$