# How do you differentiate f(x)=(2x+3)^4 / x using the chain rule?

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left({\left(2 x + 3\right)}^{4} / x\right)$
Or, $f ' \left(x\right) = \frac{1}{x} \cdot \frac{d}{d \left(2 x + 3\right)} {\left(2 x + 3\right)}^{4} \cdot \frac{d}{\mathrm{dx}} \left(2 x + 3\right) + {\left(2 x + 3\right)}^{4} \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$
$= \frac{1}{x} \cdot 4 {\left(2 x + 3\right)}^{3} \cdot 2 - {\left(2 x + 3\right)}^{4} / {x}^{2}$
$= \frac{8 x {\left(2 x + 3\right)}^{3} - {\left(2 x + 3\right)}^{4}}{x} ^ 2$
$= {\left(2 x + 3\right)}^{3} \frac{8 x - 2 x - 3}{x} ^ 2$
$= 3 {\left(2 x + 3\right)}^{3} \cdot \frac{2 x - 1}{x} ^ 2$