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

Apr 11, 2018

$\frac{3}{2} \cdot \left(\sqrt{{x}^{3} - 2 x + 3}\right) \cdot \left(3 {x}^{2} - 2\right)$

#### Explanation:

The chain rule:
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

The power rule:
$\frac{d}{\mathrm{dx}} {x}^{n} = n \cdot {x}^{n - 1}$

Applying these rules:
1 The inner function, $g \left(x\right)$ is ${x}^{3} - 2 x + 3$, the outer function, $f \left(x\right)$ is $g {\left(x\right)}^{\frac{3}{2}}$

2 Take the derivative of the outer function using the power rule
$\frac{d}{\mathrm{dx}} {\left(g \left(x\right)\right)}^{\frac{3}{2}} = \frac{3}{2} \cdot g {\left(x\right)}^{\frac{3}{2} - \frac{2}{2}} = \frac{3}{2} \cdot g {\left(x\right)}^{\frac{1}{2}} = \frac{3}{2} \cdot \sqrt{g \left(x\right)}$
$f ' \left(g \left(x\right)\right) = \frac{3}{2} \cdot \sqrt{{x}^{3} - 2 x + 3}$

3 Take the derivative of the inner function
$\frac{d}{\mathrm{dx}} g \left(x\right) = 3 {x}^{2} - 2$
$g ' \left(x\right) = 3 {x}^{2} - 2$

4 Multiply $f ' \left(g \left(x\right)\right)$ with $g ' \left(x\right)$
$\left(\frac{3}{2} \cdot \sqrt{{x}^{3} - 2 x + 3}\right) \cdot \left(3 {x}^{2} - 2\right)$

solution: $\frac{3}{2} \cdot \left(\sqrt{{x}^{3} - 2 x + 3}\right) \cdot \left(3 {x}^{2} - 2\right)$