What is the derivative of f(x) = x^2(x-2)^4?

Apr 6, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 x {\left(x - 2\right)}^{3} \left(3 x - 2\right)$

Explanation:

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

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

Here $g \left(x\right) = {x}^{2}$ and $h \left(x\right) = {\left(x - 2\right)}^{4}$

Hence, $\frac{\mathrm{df}}{\mathrm{dx}} = {x}^{2} \times 4 {\left(x - 2\right)}^{3} + {\left(x - 2\right)}^{4} \times 2 x$

= $2 x {\left(x - 2\right)}^{3} \left\{2 x + \left(x - 2\right)\right\}$

= $2 x {\left(x - 2\right)}^{3} \left(3 x - 2\right)$