Question

How do you differentiate `f(x)= (x^2 + 2x )/( x^2 -4)`?

Answer 1

`f'(x)=-2/(x-2)^2`

Explanation:

Since the given function is in the form of a fraction, we can use the quotient rule.

The quotient rule states that,

`color(blue)(|bar(ul(color(white)(a/a)color(black)((u/v)'=(vu'-uv')/(v^2))color(white)(a/a)|)))`

In your case:

Let `x^2+2x` be `u`.
Let `x^2-4` be `v`.

Plugging in the values into the quotient rule,

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

Use the power rule to differentiate `(x^2+2x)` and `(x^2-4)`.

Note: Use the formula, `(x^n)'=n*x^(n-1)`.

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

Simplifying,

`f'(x)=((2x^3+2x^2-8x-8)-(2x^3+4x^2))/((x+2)^2(x-2)^2)`

Removing the brackets in the numerator,

`f'(x)=(2x^3+2x^2-8x-8-2x^3-4x^2)/((x+2)^2(x-2)^2)`

`f'(x)=(color(red)cancelcolor(black)(2x^3)+2x^2-8x-8color(red)cancelcolor(black)(-2x^3)-4x^2)/((x+2)^2(x-2)^2)`

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

Factor out `-2` from the numerator.

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

Factor the leftover expression in the numerator.

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

`f'(x)=(-2color(red)cancelcolor(black)((x+2)^2))/(color(red)cancelcolor(black)((x+2)^2)(x-2)^2)`

`f'(x)=color(green)(|bar(ul(color(white)(a/a)color(black)(-2/(x-2)^2)color(white)(a/a)|)))`