# How do you find the derivative of  y=x^3(7x-7)^5 using the chain rule?

Feb 20, 2016

$y ' = \left(56 {x}^{3} - 21 {x}^{2}\right) {\left(7 x - 7\right)}^{4}$

#### Explanation:

We'll have to use a little more than the chain rule to get our result.

${x}^{3}$ and ${\left(7 x - 7\right)}^{5}$ are being multiplied by each other, so we have to use the product rule, which states
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
Where $u$ and $v$ are functions of $x$. Let's have $u = {x}^{3}$ and $v = {\left(7 x - 7\right)}^{5}$. Now we differentiate them:
$u = {x}^{3} \to u ' = 3 {x}^{2}$

$v = {\left(7 x - 7\right)}^{5} \to v ' = 5 \cdot 7 {\left(7 x - 7\right)}^{4} = 35 {\left(7 x - 7\right)}^{4}$
In differentiating $v$, we have to use the chain rule - meaning we find the derivative of the "inside" function $\left(7 x - 7\right)$ and multiply it by the "outside" function $\left({\left(7 x - 7\right)}^{5}\right)$. The derivative of $7 x - 7$ is $7$, and $7$ multiplied by ${\left(7 x - 7\right)}^{5}$ is $7 {\left(7 x - 7\right)}^{5}$. To finish off, use the power rule to get $35 {\left(7 x - 7\right)}^{4}$.

Plugging everything we just found into the product rule formula, we find
$y ' = \left({x}^{3}\right) ' {\left(7 x - 7\right)}^{5} + \left({x}^{3}\right) \left({\left(7 x - 7\right)}^{5}\right) '$
$y ' = 3 {x}^{2} {\left(7 x - 7\right)}^{5} + 35 {x}^{3} {\left(7 x - 7\right)}^{4}$

If we want to, we can factor out an ${x}^{2} {\left(7 x - 7\right)}^{4}$ to simplify the result to:
$y ' = {x}^{2} {\left(7 x - 7\right)}^{4} \left(3 \left(7 x - 7\right) + 35 x\right)$
$y ' = {x}^{2} {\left(7 x - 7\right)}^{4} \left(56 x - 21\right)$

We can go even further to get:
$y ' = \left(56 {x}^{3} - 21 {x}^{2}\right) {\left(7 x - 7\right)}^{4}$