# How do you differentiate f(x)=(x^4 - 2x^2)^6 using the chain rule?

##### 1 Answer
Oct 30, 2015

$24 x {\left({x}^{4} - 2 {x}^{2}\right)}^{5} \cdot \left({x}^{2} - 1\right)$

#### Explanation:

The chain rule is used to differentiate composite functions, and your two functions are

$\left\{\begin{matrix}f \left(x\right) = {x}^{6} \\ g \left(x\right) = {x}^{4} - 2 {x}^{2}\end{matrix}\right.$

And your composition is $f \left(g \left(x\right)\right)$.

The chain rule tells you that the derivative of $f \left(g \left(x\right)\right)$ is $f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$, and we have that

$\left\{\begin{matrix}f ' \left(x\right) = 6 {x}^{5} \\ g ' \left(x\right) = 4 {x}^{3} - 4 x\end{matrix}\right.$

So:

$f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = 6 {\left({x}^{4} - 2 {x}^{2}\right)}^{5} \cdot \left(4 {x}^{3} - 4 x\right)$.

Since we can factor $4 x$ from the last factor, we can rewrite it as

$24 x {\left({x}^{4} - 2 {x}^{2}\right)}^{5} \cdot \left({x}^{2} - 1\right)$