# How do you use the chain rule to differentiate y=root3(-2x^4+5)?

Jun 12, 2017

y'=(-8x^3)/(3root3((-2x+5)^2) Or alternatively y'=-(8x^3)/(3root3((5-2x)^2

#### Explanation:

The first step is to rewrite the equation using powers:

$y = {\left(- 2 {x}^{4} + 5\right)}^{\frac{1}{3}}$

Now we are able to apply the chain rule, we basically take the derivative of the outside times the derivative of the inside. You will need the power rule too.

$\frac{d}{\mathrm{dx}} = \frac{1}{3} {\left(- 2 {x}^{4} + 5\right)}^{- \frac{2}{3}} \times \frac{d}{\mathrm{dx}} \left(- 2 {x}^{4} + 5\right)$

$\frac{d}{\mathrm{dx}} = \textcolor{b l u e}{\frac{1}{3} {\left(- 2 {x}^{4} + 5\right)}^{- \frac{2}{3}}} \times \left(- 8 {x}^{3}\right)$

What we want to do now is rewrite what's in blue:

color(blue)(1/3(-2x^4+5)^(-2/3))=1/(3root3((-2x+5)^2)

Now that we know this we can simply multiply straight through:

$\frac{d}{\mathrm{dx}} = \frac{1}{3 \sqrt{{\left(- 2 x + 5\right)}^{2}}} \times \frac{- 8 {x}^{3}}{1}$

y'=(-8x^3)/(3root3((-2x+5)^2) Or y'=-(8x^3)/(3root3((5-2x)^2