How do you find the derivative of $h (x) = {(6 x - x^{3})}^{2}$ $h(x) = (6x-x^3)^2$ ?

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Answer 1 · 2016-06-14T18:12:14

$y = u^{2}$ $y = u^2$

$u = (6 x - x^{3})$ $u = (6x - x^3)$

The chain rule states that $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $dy/dx = dy/(du) xx (du)/dx$ .

We must therefore differentiate both functions. Starting with $y$ $y$ :

$y' = 2u^(2 - 1)$

$y' = 2u$

Now for $u$ :

$u' = 6 - 3x^2$

$:. dy/dx = 2u xx (6 - 3x^2)$

$dy/dx = 2(6x - x^3) xx (6 - 3x^2)$

$dy/dx = (12x - 2x^3)(6 - 3x^2)$

$dy/dx = 72x - 12x^3 + 6x^5 - 36x^3$

$dy/dx = 6x^5 - 48x^3 + 72x$

Hopefully this helps!

How do you find the derivative of h(x) = (6x-x^3)^2h(x)=(6x−x3)2h(x) = (6x-x^3)^2?