AP Calculus AB and BC

Derivative of x/(x+1): Answer, Proof, Mistakes

The derivative of x/(x+1) with respect to x is 1/(x+1)^2. The quotient rule gives [(x+1) times 1 minus x times 1] over (x+1)^2, and the numerator collapses to (x+1) minus x = 1, leaving 1/(x+1)^2. Because the numerator is 1 and the denominator is a square, the slope is positive everywhere except at x = -1.

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

The proof: quotient rule on x/(x+1)

With top g=xg = x and bottom h=x+1h = x + 1, the quotient rule is g′h−gh′h2\frac{g'h - gh'}{h^2}. Here g′=1g' = 1 and h′=1h' = 1.

ddx[xx+1]=(x+1)⋅1−x⋅1(x+1)2=x+1−x(x+1)2=1(x+1)2\frac{d}{dx}\left[\frac{x}{x+1}\right] = \frac{(x+1)\cdot 1 - x\cdot 1}{(x+1)^2} = \frac{x + 1 - x}{(x+1)^2} = \frac{1}{(x+1)^2}

The numerator simplifies to x+1−x=1x + 1 - x = 1, so the whole derivative reduces to 1(x+1)2\frac{1}{(x+1)^2}. The denominator stays squared, which is what keeps the slope positive.

A faster route: rewrite the fraction

You can avoid the quotient rule entirely. Split the fraction as xx+1=(x+1)−1x+1=1−1x+1\frac{x}{x+1} = \frac{(x+1) - 1}{x+1} = 1 - \frac{1}{x+1}, then differentiate term by term.

ddx[1−(x+1)−1]=0−(−(x+1)−2)=1(x+1)2\frac{d}{dx}\left[1 - (x+1)^{-1}\right] = 0 - \left(-(x+1)^{-2}\right) = \frac{1}{(x+1)^2}

Same answer, and it shows why the derivative is always positive: it is the derivative of −1x+1-\frac{1}{x+1}.

Common mistakes

  • Flipping the order of the quotient rule to gh′−g′hh2\frac{gh' - g'h}{h^2}, which gives −1(x+1)2-\frac{1}{(x+1)^2}. The correct order is g′h−gh′g'h - gh', low times derivative of high minus high times derivative of low.
  • Forgetting to square the denominator and writing 1x+1\frac{1}{x+1}. The quotient rule always divides by h2h^2.
  • Differentiating numerator over denominator separately as 11=1\frac{1}{1} = 1. You cannot differentiate top and bottom independently; that is not a valid rule.
  • Canceling the xx in xx+1\frac{x}{x+1} to get 11\frac{1}{1} before differentiating. The xx and x+1x+1 share no common factor.

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/(x+1)?

It is 1(x+1)2\frac{1}{(x+1)^2}. The quotient rule numerator (x+1)−x(x+1) - x collapses to 11, leaving 1(x+1)2\frac{1}{(x+1)^2}.

Why is the derivative always positive?

The numerator is 11 and the denominator (x+1)2(x+1)^2 is a square, so the ratio is positive for every x≠−1x \neq -1. The function increases on each side of the asymptote at x=−1x = -1.

Can I find this derivative without the quotient rule?

Yes. Rewrite xx+1=1−1x+1\frac{x}{x+1} = 1 - \frac{1}{x+1} and differentiate to get 1(x+1)2\frac{1}{(x+1)^2}, the same result.