# How do you find the derivative of y=(3x^5 - 4x^3 + 2)^23 using the chain rule?

Dec 24, 2015

$y ' = 69 {x}^{2} {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \left(5 {x}^{2} - 4\right)$

#### Explanation:

According to the chain rule,

$\frac{d}{\mathrm{dx}} \left[{u}^{23}\right] = 23 {u}^{22} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

$y ' = 23 {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \cdot \frac{d}{\mathrm{dx}} \left[3 {x}^{5} - 4 {x}^{3} + 2\right]$

$\implies 23 {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \left(15 {x}^{4} - 12 {x}^{2}\right)$

Factor out a $3 {x}^{2}$ from the last term.

$\implies 69 {x}^{2} {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \left(5 {x}^{2} - 4\right)$