Question

How do you find the derivative of `g(x)= x^2sqrt(1-x^2)`?

Answer 1

`g^' = (x * (2 - 3x^2))/(sqrt(1-x^2))`

Explanation:

You can use the product rule and the chain rule to differentiate this function, which can be rewritten as

`g(x) = x^2 * (1-x^2)^(1/2)`

The product rule allows you to differentiate functions that take the form

`color(blue)(g(x) = f(x) * h(x))`

by using the formula

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

In your case, you have

`f(x) = x^2" "` and `" "h(x) = (1-x^2)^(1/2)`

This means that you can write

`d/dx(g(x)) = [d/dx(x^2)] * (1-x^2)^(1/2) + x^2 * d/dx(1-x^2)^(1/2)`

To find `d/dx(h(x))`, you need to use the chain rule for `u^(1/2)`, with `u = (1-x^2)`. This will get you

`d/dx(u^(1/2)) = d/(du)u^(1/2) * d/dx(u)`

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

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

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

Plug this into your target derivative to get

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

This can be simplified to get

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

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

Finally, you can rewrite this as

`g^' = color(green)((x * (2 - 3x^2))/(sqrt(1-x^2)))`