Derivative of (x^2+1)/(x−1) by the Quotient Rule
The problem: differentiate f(x) = (x² + 1)/(x − 1) using the quotient rule.
Students find the quotient rule hard for one reason above all: the minus sign in the numerator. The formula (f′g − fg′)/g² looks innocent on a formula sheet, but under exam pressure that minus becomes a plus, or it fails to distribute over the whole second term, and the answer silently flips. This example works the derivative of (x²+1)/(x−1) slowly enough that the minus sign cannot hide, then verifies the result numerically so you know it is right.
Worked solution
We want f′(x) for f(x) = (x² + 1)/(x − 1).
Step 1. Name the top and bottom. Let the numerator be N = x² + 1 and the denominator be D = x − 1. The quotient rule says:
(N/D)′ = (N′D − ND′)/D²
Say it in words once: derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.
Step 2. Differentiate the pieces. N′ = 2x and D′ = 1. Nothing fancy; do these first and write them down so they are ready.
Step 3. Plug into the formula. This is where care pays:
f′(x) = [2x·(x − 1) − (x² + 1)·1] / (x − 1)²
Notice the parentheses around (x² + 1): the minus sign in front of it must eventually distribute over the entire numerator piece, not just the x².
Step 4. Expand the numerator. Multiply out each part separately:
2x(x − 1) = 2x² − 2x
(x² + 1)(1) = x² + 1
Step 5. Subtract carefully. This is the #1 error step in all of the quotient rule, so go slowly:
(2x² − 2x) − (x² + 1) = 2x² − 2x − x² − 1
The minus distributes to both terms inside the parentheses: minus x² and minus 1. If you wrote 2x² − 2x − x² + 1 instead, you have just committed the classic sign error.
Step 6. Combine like terms. 2x² − x² = x², and −2x and −1 have no partners:
f′(x) = (x² − 2x − 1) / (x − 1)², for x ≠ 1
Step 7. Note the domain. The original function is undefined at x = 1 (division by zero), so the derivative is undefined there too. The formula’s (x − 1)² denominator enforces this automatically, but state it: x ≠ 1.
Why this method works
The quotient rule is the product rule in disguise: N/D = N · D⁻¹, and differentiating that product with the chain rule produces exactly (N′D − ND′)/D². The minus sign is not arbitrary; it comes from the derivative of D⁻¹, which is −D⁻² · D′. Understanding the origin helps under pressure: the minus belongs to the second term, the whole second term, every time.
Common mistakes
- The minus that is not distributed. Writing (2x² − 2x) − (x² + 1) as 2x² − 2x − x² + 1. The parentheses after a minus sign mean “subtract everything inside”: both x² and 1 change sign.
- A plus instead of a minus. Writing (N′D + ND′)/D² confuses the quotient rule with the product rule. Mnemonic: quotients involve division, and the formula “divides” the two terms with a minus.
- Forgetting to square the denominator. Writing (x − 1) instead of (x − 1)² at the end. The D² comes from the derivation and is not optional; without it the units and the values are both wrong.
Check (never skip)
Test at x = 2. Our formula gives f′(2) = (4 − 4 − 1)/(2 − 1)² = −1/1 = −1. Now approximate numerically from the original function: f(2) = (4 + 1)/(2 − 1) = 5, and f(2.001) = (4.004001 + 1)/(1.001) = 5.004001/1.001 ≈ 4.999002. Then [f(2.001) − f(2)]/0.001 ≈ (4.999002 − 5)/0.001 = −0.998, which matches −1 to the accuracy of a 0.001 step. ✓ The quotient rule output agrees with the raw definition of the derivative, so (x² − 2x − 1)/(x − 1)² is correct.
Frequently Asked Questions
Q: When do I use the quotient rule instead of the product rule?
A: Use the quotient rule for one function divided by another. You can equivalently write N/D as N · D⁻¹ and use the product rule plus the chain rule; the algebra is longer but the answer is identical.
Q: What is the single most common quotient-rule mistake?
A: The minus sign in (N′D − ND′). Either it becomes a plus, or it fails to distribute across the whole ND′ term. Circling the minus before you expand is a cheap habit that prevents an expensive error.
Q: Why is x = 1 excluded from the answer?
A: Because the denominator (x − 1)² equals zero at x = 1, and the original function (x²+1)/(x−1) is undefined there too. A derivative cannot exist where the function itself does not.
Try it with steps on our derivative calculator.