Question

How do you differentiate `f(x)=(x^2+4x)*(2+x^-2)` using the product rule?

Answer 1

`f'(x) = 4x + 8 - 4/x^2`

Explanation:

`f'(x) = frac{"d"}{"d"x}((x^2+4x)*(2+x^{-2}))`

`= (x^2+4x)*frac{"d"}{"d"x}(2+x^{-2})`

`+ (2+x^{-2})*frac{"d"}{"d"x}(x^2+4x)`

`= (x^2+4x)*((-2)x^{-3})`

`+ (2+x^{-2})*(2x+4)`

`= (-2/x-8/x^2) + (4x +2/x + 8 + 4/x^2)`

`= 4x + 8 - 4/x^2`

Answer 2

`-2x^-2(x+4) + 2(2+x^-2)(x+2)`

Explanation:

using the `color(blue)(" product rule ")`

If f(x) = g(x).h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)

hence `f'(x) = (x^2+4x) d/dx(2+x^-2) + (2+x^-2) d/dx(x^2+4x)`

`= (x^2+4x)(-2x^-3) + (2+x^-2)(2x+4)`

`= x(x+4)(-2x^-3) + (2+x^-2) 2(x+2)`
(removed 'common factors' from the 2 brackets of x and 2)

`rArr = -2x^-2(x+4) + 2(2+x^-2)(x+2)`