Question

How do you differentiate `(2x)/(x+1)^2`?

Answer 1

`(f/g)'=(-color(red)(2)(x-1))/(x+1)^3`

Explanation:

The quotient rule tells us that for a function in the form of `f/g, (f/g)'=(gf'-fg')g^2`

Here, `g(x)=(x+1)^2`

`g'(x)=d/dx(x^2+2x+1)=2x+2`

`f(x)=2x`

`f'(x)=2`

So,

`(f/g)'=(2(x+1)^2-(2x+2)(2x))/(x+1)^4`

Simplify.

`(f/g)'=(2(x^2+2x+color(red)(1))-(4x^2+4x))/(x+1)^4`

`(f/g)'=(2x^2+4x+color(red)(2)-4x^2-4x)/(x+1)^4`

`(f/g)'=color(red)((2-2x^2))/(x+1)^4`

`(f/g)'=(-color(red)(2)(x^2-1))/(x+1)^4`

`(f/g)'=(-color(red)(2)cancel((x+1))(x-1))/(x+1)^((cancel4)3)`

`(f/g)'=(-color(red)(2)(x-1))/(x+1)^3`

Answer 2

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

Explanation:

`f(x) = (2x) / (x+1)^2`

`u = 2x, v = (x+1)^2`

`du = 2, dv = 2 (x+1) * 1 = 2(x+1)`

Using quotient rule,

`f'(x) = (v du - u dv) / v^2`

`f'(x) =( ((x+1)^2 * 2) - (2x * (2(x+1)))) / (x+1)^4`

`f'(x) = (2x^2 + 4x + 2 - 4x^2 - 4x) / (x+1)^4`

`f'(x) =( 2 - 2x^2) / (x+1)^4`

`f'(x) = (2 cancel(color(brown) (1 + x)) * (1-x) ) / (x+1)^cancel(color(brown)4^color(green)(3)`#

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