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

1 Answer
Aug 29, 2015

Answer:

#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)#