AP Calculus AB and BC

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

The derivative of (x^2-1)/(x^2+1) is 4x/(x^2+1)^2. The quotient rule decides everything: subtracting the second product flips the sign of its 2x term, so the two cross terms add to 4x instead of cancelling. The only critical point is x = 0, and it is a minimum.

ddx[x2−1x2+1]=4x(x2+1)2\frac{d}{dx}\left[\frac{x^{2}-1}{x^{2}+1}\right] = \frac{4x}{\left(x^{2}+1\right)^{2}}

The quotient rule, term by term

Set u=x2−1u = x^{2}-1 and v=x2+1v = x^{2}+1, so u′=2xu' = 2x and v′=2xv' = 2x. The two derivatives are identical, and the whole character of the answer depends on what the subtraction does to them.

f′(x)=2x(x2+1)−(x2−1)⋅2x(x2+1)2f'(x) = \frac{2x\left(x^{2}+1\right)-\left(x^{2}-1\right)\cdot 2x}{\left(x^{2}+1\right)^{2}}

Expand carefully: 2x3+2x−(2x3−2x)=2x3+2x−2x3+2x2x^{3}+2x-\left(2x^{3}-2x\right) = 2x^{3}+2x-2x^{3}+2x. The cubic terms cancel, but the two linear terms both end up positive and add.

f′(x)=4x(x2+1)2f'(x) = \frac{4x}{\left(x^{2}+1\right)^{2}}

What 4x over (x^2+1)^2 says about the graph

The denominator is positive for every xx, so the sign of f′f' is just the sign of xx: negative to the left of 00, positive to the right. That makes x=0x=0 the only critical point and a minimum, with f(0)=−1f(0) = -1.

For the ends, divide top and bottom of ff by x2x^{2}: 1−1x21+1x2→1\frac{1-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}\to 1 in both directions. So y=1y=1 is a horizontal asymptote on the left and the right, and the graph dips from near 11 down to −1-1 at the origin before climbing back.

A second route confirms the answer. Long division gives f(x)=1−2x2+1f(x) = 1-\frac{2}{x^{2}+1}, and differentiating −2(x2+1)−1-2\left(x^{2}+1\right)^{-1} by the chain rule gives 2(x2+1)−2⋅2x=4x(x2+1)22\left(x^{2}+1\right)^{-2}\cdot 2x = \frac{4x}{\left(x^{2}+1\right)^{2}}. The rewrite also makes the asymptote obvious.

The mistakes students make

The first two errors are sign errors inside the numerator; the third mixes up zeros of f with zeros of f'.

  • Answering 2x(x2+1)2\frac{2x}{\left(x^{2}+1\right)^{2}} after factoring the numerator as 2x[(x2+1)−(x2−1)]2x\left[\left(x^{2}+1\right)-\left(x^{2}-1\right)\right] and reading the bracket as 11. The x2x^{2} terms cancel there, but the −(−1)-\left(-1\right) is a second +1+1, so the bracket is 22 and the numerator is 4x4x.
  • Dropping the minus and computing 2x(x2+1)+(x2−1)2x2x\left(x^{2}+1\right)+\left(x^{2}-1\right)2x, which gives 4x3(x2+1)2\frac{4x^{3}}{\left(x^{2}+1\right)^{2}}. The cubes only cancel when the second product is subtracted.
  • Reporting critical points at x=±1x=\pm 1 because f(±1)=0f\left(\pm 1\right)=0. Those are the xx-intercepts of the function, not zeros of the derivative. The derivative is zero only at x=0x=0.

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^2-1)/(x^2+1)?

It is 4x(x2+1)2\frac{4x}{\left(x^{2}+1\right)^{2}}.

Why is the numerator 4x and not 2x?

The quotient rule numerator is 2x(x2+1)−(x2−1)2x=2x3+2x−2x3+2x2x\left(x^{2}+1\right)-\left(x^{2}-1\right)2x = 2x^{3}+2x-2x^{3}+2x. Subtracting −2x-2x adds it, so the two linear terms combine into 4x4x.

Does (x^2-1)/(x^2+1) have a maximum or minimum?

It has a minimum of −1-1 at x=0x=0 and no maximum. The values climb toward the horizontal asymptote y=1y=1 in both directions without ever reaching it.