How do you differentiate $\frac{3}{4} \cdot {(2 x^{3} + 3 x)}^{- \frac{1}{4}}$ $3/4 * (2x^3 + 3x)^(-1/4)$ ?

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Answer 1 · 2016-06-22T15:27:46

$\frac{- 18 x^{2} + 9}{16 {(2 x^{3} + 3 x)}^{\frac{5}{4}}}$ $(-18x^2+9)/(16(2x^3+3x)^(5/4)$

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

Then we apply the chain rule: $d/dx[f(g(x))] -> f'(g(x)) * g'(x)$

$d/dx[(2x^3+3x)^(-1/4)] -> -1/4(2x^3+3x)^(-5/4)$

$d/dx[2x^3+3x] -> 6x^2+3$

We multiply the outside derivative and inside derivative together:
$3/4 * [-1/4(2x^3+3x)^(-5/4) * (6x^2+3)]$

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

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

How do you differentiate 3/4 * (2x^3 + 3x)^(-1/4)34⋅(2x3+3x)−14 3/4 * (2x^3 + 3x)^(-1/4)?