# How do you differentiate 3sin^5(2x^2)  using the chain rule?

Dec 25, 2015

Break the given chain rule problem in manageable links and use it to differentiate. Explanation is given below.

#### Explanation:

The Chain rule is easy if you split the chain into manageable links.

Let us see how we can do this with respect to our problem

$3 {\sin}^{5} \left(2 {x}^{2}\right)$
$y = 3 {\sin}^{5} \left(2 {x}^{2}\right)$

We can split the links as below
$y = 3 {u}^{5}$
$u = \sin \left(v\right)$
v=2x^

The chain rule is given by the following
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$

Note how each link is differentiated separately and multiplied to form the chain.

$y = 3 {u}^{5}$
$\frac{\mathrm{dy}}{\mathrm{du}} = 15 {u}^{4}$

$u = \sin \left(v\right)$
$\frac{\mathrm{du}}{\mathrm{dv}} = \cos \left(v\right)$

$v = 2 {x}^{2}$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 4 x$

Applying chain rule we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = 15 {u}^{4} \cdot \cos \left(v\right) \cdot 4 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 60 x {u}^{4} \cos \left(v\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 60 x {\sin}^{4} \left(v\right) \cos \left(v\right)$ substituted $u = \sin \left(v\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 60 x {\sin}^{4} \left(2 {x}^{2}\right) \cos \left(2 {x}^{2}\right)$ substituted $v = 2 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 60 x {\sin}^{4} \left(2 {x}^{2}\right) \cos \left(2 {x}^{2}\right)$