# How do you differentiate  f(x)=(7-8x^3)^2 using the chain rule.?

Apr 1, 2016

$f ' \left(x\right) = 384 {x}^{5} - 336 {x}^{2}$

#### Explanation:

If $f \left(x\right) = g \left(h \left(x\right)\right)$ then $f ' \left(x\right) = h ' \left(x\right) \cdot g ' \left(h \left(x\right)\right)$.

In other words, if the function you have can be looked at as one function inside another, then the derivative of the whole thing is the derivative of the inside multiplied by the derivative of the outside with the inside left the same.

It sounds harder than it is.

We can look at $f \left(x\right) = {\left(7 - 8 {x}^{3}\right)}^{2}$ as an inside function, $h \left(x\right) = 7 - 8 {x}^{3}$, and and outside function, $g \left(h\right) = h {\left(x\right)}^{2}$.

Therefore, to differentiate the whole thing you do $h ' \left(x\right) \cdot g ' \left(h\right)$, like so:

$h ' \left(x\right) = 0 - 24 {x}^{2}$
$= - 24 {x}^{2}$
$g ' \left(h\right) = 2 \cdot h \left(x\right)$
$= 2 \left(7 - 8 {x}^{3}\right)$
$= 14 - 16 {x}^{3}$

$h ' \left(x\right) g ' \left(h \left(x\right)\right) = - 24 {x}^{2} \cdot \left(14 - 16 {x}^{3}\right)$
$= - 336 {x}^{2} + 384 {x}^{5}$

You can also solve this problem by expanding out the original brackets and differentiating in the usual fashion, and should (hopefully) arrive at the same answer.

In summary, chain rule is $\frac{\mathrm{dy}}{\mathrm{dx}}$ inside $\cdot$ $\frac{\mathrm{dy}}{\mathrm{dx}}$ outside, with the inside untouched.