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

Feb 5, 2016

See the explanation below.

#### Explanation:

If we leave it as written , we have:

${\left({\sin}^{3} \left(2 {x}^{2}\right)\right)}^{\frac{1}{2}}$

So the derivative is

1/2 (sin^3(2x^2) )^(-1/2)[d/dx(sin^3(2x^2)]

$= \frac{1}{2} {\left({\sin}^{3} \left(2 {x}^{2}\right)\right)}^{- \frac{1}{2}} \left[3 {\sin}^{2} \left(2 {x}^{2}\right) \left(\frac{d}{\mathrm{dx}} \left(\sin \left(2 {x}^{2}\right)\right)\right)\right]$

$= \frac{1}{2} {\left({\sin}^{3} \left(2 {x}^{2}\right)\right)}^{- \frac{1}{2}} \left[3 {\sin}^{2} \left(2 {x}^{2}\right) \cos \left(2 {x}^{2}\right) \left(\frac{d}{\mathrm{dx}} \left(2 {x}^{2}\right)\right)\right]$

$= \frac{1}{2} {\left({\sin}^{3} \left(2 {x}^{2}\right)\right)}^{- \frac{1}{2}} \left[3 {\sin}^{2} \left(2 {x}^{2}\right) \cos \left(2 {x}^{2}\right) \left(4 x\right)\right]$

Now simplify.

If we rewrite as

${\left(\sin \left(2 {x}^{2}\right)\right)}^{\frac{3}{2}}$

We get

$\frac{3}{2} {\left(\sin \left(2 {x}^{2}\right)\right)}^{\frac{1}{2}} \cos \left(2 {x}^{2}\right) \left(4 x\right)$

$= 6 x \cos \left(2 {x}^{2}\right) \sqrt{\sin \left(2 {x}^{2}\right)}$