# How do you differentiate f(x) =x^3sqrt(2x+1)  using the product rule?

Jun 23, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = 3 {x}^{2} \sqrt{2 x + 1} + {x}^{3} / \sqrt{2 x + 1}$

#### Explanation:

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

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

Hence, as $f \left(x\right) = {x}^{3} \sqrt{2 x + 1}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = 3 {x}^{2} \times \sqrt{2 x + 1} + \frac{1}{2} {\left(2 x + 1\right)}^{- \frac{1}{2}} \times 2 \times {x}^{3}$

i.e. $\frac{\mathrm{df}}{\mathrm{dx}} = 3 {x}^{2} \sqrt{2 x + 1} + {x}^{3} / \sqrt{2 x + 1}$