Question

How do you find the derivative of `-2x(x^2+3)^(-2)`?

Answer 1

`y^' = (6(x-1)(x+1))/(x^2 + 3)^3`

Explanation:

One way in which you can differentiate this function is by using the product rule and the chain rule.

You know that for a function `f(x)` that can be written as the product of two other functions, `g(x)` and `h(x)`, its derivative can be calculated using

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

In your case, you have

`g(x) = -2x" "` and `" "h(x) = (x^2 + 3)^(-2)`

This means that you have

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

To differentiate `(x^2 + 3)^(-2)`, use the chain rule for `u^(-2)`, with `u = x^2 + 3`. This will get you

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

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

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

`d/dx((x^2 + 3)^(-2)) = -4x(x^2 + 3)^(-2)`

Take this back to your target derivative

`y^' = (-2) * (x^2+3)^(-2) + (-2x) * [-4x(x^2+3)^(-3)]`

`y^' = -2*(x^2+3)^(-3) * (x^2 + 3 - 4x^2)`

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

`y^' = (6(x^2-1))/(x^2 + 3)^3 = color(green)((6(x-1)(x+1))/(x^2 + 3)^3)`