AP Calculus AB and BC

Derivative of x^2/(x+1): Quotient Rule

The derivative of x squared over x plus 1 is x squared plus 2x, all over x plus 1 squared. The numerator factors as x times x plus 2, so the critical points are at x equals 0 and x equals negative 2, with a vertical asymptote at negative 1 in between.

ddx[x2x+1]=x2+2x(x+1)2\frac{d}{dx}\left[\frac{x^{2}}{x+1}\right] = \frac{x^{2}+2x}{\left(x+1\right)^{2}}

The quotient rule working

Take u=x2u = x^{2} and v=x+1v = x+1, so u′=2xu' = 2x and v′=1v' = 1.

2x(x+1)−x2(1)(x+1)2=2x2+2x−x2(x+1)2=x2+2x(x+1)2\frac{2x(x+1) - x^{2}(1)}{(x+1)^{2}} = \frac{2x^{2}+2x-x^{2}}{(x+1)^{2}} = \frac{x^{2}+2x}{(x+1)^{2}}

Factoring the numerator as x(x+2)x(x+2) makes the critical points obvious.

The asymptote sits between the critical points

The function has a vertical asymptote at x=−1x = -1, and the two critical points at x=−2x = -2 and x=0x = 0 sit on either side of it. Because the asymptote separates them, one is a LOCAL maximum and the other a local minimum, but neither is a global extremum.

Long division also gives a slant asymptote: x2x+1=x−1+1x+1\frac{x^{2}}{x+1} = x - 1 + \frac{1}{x+1}, so the graph approaches y=x−1y = x-1 at both ends.

Common mistakes

  • Cancelling the xx terms. Neither is a factor of the whole expression.
  • Missing x=−2x = -2 because the factored form was never written down.
  • Treating x=−1x = -1 as a critical point. The function is undefined there, so it is an asymptote, not a critical point.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of x^2/(x+1)?

It is x2+2x(x+1)2\frac{x^{2}+2x}{(x+1)^{2}}, or x(x+2)(x+1)2\frac{x(x+2)}{(x+1)^{2}} factored.

Where are the critical points?

At x=0x = 0 and x=−2x = -2, the zeros of the numerator.

Does it have a slant asymptote?

Yes, y=x−1y = x - 1, from the long division x2x+1=x−1+1x+1\frac{x^{2}}{x+1} = x - 1 + \frac{1}{x+1}.