# Question #c7514

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {\sin}^{4} x \cos x$
${\sin}^{5} x$ is a shorthand for ${\left(\sin x\right)}^{5}$. There are two functions involved here: $g \left(x\right) = \sin x$ and $f \left(x\right) = {x}^{5}$, composed such that $y = f \left(g \left(x\right)\right)$, which can be differentiated using the chain rule.
The chain rule states that $y ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$. $f ' \left(x\right) = 5 {x}^{4}$ and $g ' \left(x\right) = \cos x$. Therefore, $y ' = 5 {\left(\sin x\right)}^{4} \cdot \cos x = 5 {\sin}^{4} x \cos x$.
Alternatively, if you want you can do this with no intermediate steps by using the chain rule in the form that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ for any function $u \left(x\right)$. You can say that $\frac{d \left({\left(\sin x\right)}^{5}\right)}{\mathrm{dx}} = \frac{d \left({\left(\sin x\right)}^{5}\right)}{d \left(\sin x\right)} \cdot \frac{d \left(\sin x\right)}{\mathrm{dx}} = 5 {\left(\sin x\right)}^{4} \cdot \cos x = 5 {\sin}^{4} x \cos x$