# How do you differentiate  3/4 * (2x^3 + 3x)^(-1/4)?

Jun 22, 2016

(-18x^2+9)/(16(2x^3+3x)^(5/4)

#### Explanation:

We can factor a constant out of the equation:
$\frac{3}{4} \cdot \left[{\left(2 {x}^{3} + 3 x\right)}^{- \frac{1}{4}}\right]$

Then we apply the chain rule: $\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] \to f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left[{\left(2 {x}^{3} + 3 x\right)}^{- \frac{1}{4}}\right] \to - \frac{1}{4} {\left(2 {x}^{3} + 3 x\right)}^{- \frac{5}{4}}$

$\frac{d}{\mathrm{dx}} \left[2 {x}^{3} + 3 x\right] \to 6 {x}^{2} + 3$

We multiply the outside derivative and inside derivative together:
$\frac{3}{4} \cdot \left[- \frac{1}{4} {\left(2 {x}^{3} + 3 x\right)}^{- \frac{5}{4}} \cdot \left(6 {x}^{2} + 3\right)\right]$

Bring the negative exponent down below:
3/4*-(6x^2+3)/(4(2x^3+3x)^(5/4)

Multiply the constant across:
(-18x^2+9)/(16(2x^3+3x)^(5/4)