How do you differentiate f(x) = (3x ^2 -3x + 8) ^4 using the chain rule?

Apr 11, 2018

$4 \left(6 x - 3\right) {\left(3 {x}^{2} - 3 x + 8\right)}^{3}$

Explanation:

Let $u = 3 {x}^{2} - 3 x + 8$
Let $v = {u}^{4}$

The differential of $u$

${u}^{'} = 3 \left(2 x\right) - 3 = 6 x - 3$

The differential of $v$

${v}^{'} = 4 {u}^{3}$

You will now multiply the differentials

$\left(6 x - 3\right) \left(4 {u}^{3}\right)$

Replace $u$ with the original function

$4 \left(6 x - 3\right) {\left(3 {x}^{2} - 3 x + 8\right)}^{3}$

Apr 11, 2018

$4 \left(6 x - 3\right) {\left(3 {x}^{2} - 3 x + 8\right)}^{3}$

Explanation:

The chain rule states that,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = 3 {x}^{2} - 3 x + 8 , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = 6 x - 3$.

Then $y = {u}^{4} , \frac{\mathrm{dy}}{\mathrm{du}} = 4 {u}^{3}$.

So, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {u}^{3} \left(6 x - 3\right)$

Substituting back our original terms, we get,

$= 4 {\left(3 {x}^{2} - 3 x + 8\right)}^{3} \left(6 x - 3\right)$

Personally, I'll make this neater, and rearrange it into

$= 4 \left(6 x - 3\right) {\left(3 {x}^{2} - 3 x + 8\right)}^{3}$