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

Jul 26, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} {\left({x}^{3} - 3\right)}^{4} + 24 {x}^{5} {\left({x}^{3} - 3\right)}^{3}$

Explanation:

This problem actually requires the application of both the product rule and the chain rule.

The product rule states that
$\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

And the chain rule states that
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

We'll start by applying the product rule:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} {\left({x}^{3} - 3\right)}^{4} + 2 {x}^{3} \left(\frac{d}{\mathrm{dx}} {\left({x}^{3} - 3\right)}^{4}\right)$

Now we use the chain rule to differentiate the parenthetical term:
$= 6 {x}^{2} {\left({x}^{3} - 3\right)}^{4} + 2 {x}^{3} \left(4 {\left({x}^{3} - 3\right)}^{3} \cdot 3 {x}^{2}\right)$

We can simplify a bit to make things cleaner and combine terms:
$= 6 {x}^{2} {\left({x}^{3} - 3\right)}^{4} + 24 {x}^{5} {\left({x}^{3} - 3\right)}^{3}$