Question

How do you use the Product Rule to find the derivative of `y = x(x^2 - 2)^(3/2)`?

Answer 1

`y^' = 2 * (2x^2 - 1) * (x^2 -2 )^(1/2)`

Explanation:

Your function can be written as the product of two other functions

`y = underbrace(x)_(color(orange)(f(x))) * underbrace((x^2-2)^(3/2))_(color(red)(g(x)))`

which means that you can differentiate it by using the product rule

`color(blue)(d/dx(f(x)) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))`

In your case, the derivative of `y` will be

`d/dx(y) = [d/dx(x)] * (x^2 - 2)^(3/2) + x * d/dx(x^2-2)^(3/2)`

Now, to find `d/dx(x^2 - 2)^(3/2)`, you can use the chain rule for `u^(3/2)`, with `u = (x^2 - 2)`. This will get you

`d/dx(u^(3/2)) = d/(du)u^(3/2) * d/dx(u)`

`d/dx(u^(3/2)) = 3/2 * u^(1/2) * d/dx(x^2 - 2)`

`d/dx(x^2-2)^(3/2) = 3/color(red)(cancel(color(black)(2))) * (x^2 - 2)^(1/2) * color(red)(cancel(color(black)(2)))x`

Plug this back into your target derivative to get

`y^' = 1 * (x^2-2)^(3/2) + x * [3x * (x^2 -2 )^(1/2)]`

This is equivalent to

`y^' = (x^2 - 2)^(1/2) * ( x^2 -2 + 3x^2)`

`y^' = (x^2 -2 )^(1/2) * (4x^2 - 2)`

If you want, you can write this as

`y^' = color(green)(2 * (2x^2 - 1) * (x^2 -2 )^(1/2))`