How do you differentiate f(x)= (x^2 + 2x )/( x^2 -4)f(x)=x2+2xx24?

1 Answer
Jun 13, 2016

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)|)))