# How do you calculate the derivative for y = 3(5 - x^2)^5?

Aug 1, 2015

${y}^{'} = - 30 x \cdot {\left(5 - {x}^{2}\right)}^{4}$

#### Explanation:

You can differentiate your function, whichcan be written as

$y = 3 {u}^{5}$, with $u = 5 - {x}^{2}$

by using the chain rule

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left(y\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)}$

In your case, the derivative of $y$ would be

$\frac{d}{\mathrm{dx}} \left(3 {u}^{5}\right) = \frac{d}{\mathrm{du}} \left(3 {u}^{5}\right) \cdot \frac{d}{\mathrm{dx}} \left(5 - {x}^{2}\right)$

$\frac{d}{\mathrm{dx}} \left(3 {u}^{5}\right) = 15 {u}^{4} \cdot \left(- 2 x\right)$

$\frac{d}{\mathrm{dx}} \left(3 {\left(5 - {x}^{2}\right)}^{5}\right) = 15 {\left(5 - {x}^{2}\right)}^{4} \cdot \left(- 2 x\right)$

This is equivalent to

${y}^{'} = \textcolor{g r e e n}{- 30 x \cdot {\left(5 - {x}^{2}\right)}^{4}}$