How do you use the product Rule to find the derivative of #f(x)= (9-x)(5/(x^2 -4))#?

1 Answer
Aug 16, 2015

Answer:

#f^' = (5 * (x^2 - 18x + 4))/(x^2 - 4)^2#

Explanation:

To differentiate this function using the product rule, you have to rewrite it as a product of two functions

#f(x) = underbrace(5)_(color(orange)("constant")) *underbrace((9-x))_(color(blue)(g(x))) * underbrace((x^2 - 4)^(-1))_(color(green)(h(x)))#

The product rule tells you that you can differentiate products of two functions by using the formula

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

In your case, you would have

#d/dx(f(x)) = 5 * ([d/dx(9-x)] * (x^2 - 4)^(-1) + (9-x) * d/dx(x^2-4)^(-1))#

Now, to differentiate #(x^2 - 4)^(-1)#, you can use the chain rule for #u^(-1)#, with #u = (x^2 - 4)#. THis will get you

#d/dx(u^(-1)) = d/(du)(u^(-1)) * d/dx(u)#

#d/dx(u^(-1)) = -1 * u^(-2) * d/dx(x^2 - 4)#

#d/dx(x^2-4)^(-1) = - (x^2 - 4)^(-2) * 2x#

Take this back to your target derivative

#f^' = 5 * [ -1 * (x^2 - 4)^(-1) + (9-x) * (-2x) * (x^2 - 4)^(-2)]#

#f^' = 5 * (x^2 - 4)^(-2) * [-(x^2 - 4) - 2x * (9-x)]#

#f^' = 5 * (x^2 - 4)^(-2) * (-x^2 + 4 - 18x + 2x^2)#

#f^' = 5 * (x^2 - 4) * (x^2 - 18x + 4)#

You can write this as

#f^' = color(green)((5 * (x^2 - 18x + 4))/(x^2 - 4)^2)#