Difficulty: Intermediate | Prerequisites: Limit definition of derivative, power rule, basic differentiation
The quotient rule lets you differentiate fractions of functions, but it has a specific structure that students routinely botch. Graphing a derivative from a picture of the original function tests whether you understand what the derivative measures (slopes). Power functions f(x) = xᵃ let you explore how the choice of exponent a controls where the derivative is positive, negative, zero, or undefined.
Quotient rule
If h(x) = f(x)/g(x), then h'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]². The mnemonic is "low d-high minus high d-low, over the square of what's below." In simple terms, you cannot just differentiate the top and bottom separately.
Product rule
If h(x) = f(x)·g(x), then h'(x) = f'(x)·g(x) + f(x)·g'(x). In simple terms, each factor gets a turn being differentiated while the other stays put, and you add the two results.
Graphing the derivative
The process of sketching f'(x) by reading the slopes off the graph of f(x). Where f is rising, f' is positive; where f is falling, f' is negative; where f has a peak or trough, f' crosses zero; where f has a corner or cusp, f' does not exist.
Differentiability
A function is differentiable at a point if its derivative exists there. A function fails to be differentiable at corners (sharp turns), cusps, vertical tangent lines, and discontinuities. Think of it as: can you draw exactly one tangent line at that point? If not, the derivative does not exist there.
Power function
A function of the form f(x) = xᵃ, where a is any real number. The derivative is f'(x) = a·xᵃ⁻¹. Different values of a produce very different derivative behaviour, especially at x = 0 and x = −1.
For h(x) = f(x)/g(x):
h'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]²
Notice the structure: differentiate the numerator, multiply by the denominator, subtract the numerator times the derivative of the denominator, all over the denominator squared.
Let f(x) = x² − 5x + 1 and g(x) = 2x + 5.
f'(x) = 2x − 5
g'(x) = 2
Applying the quotient rule:
h'(x) = [(2x − 5)(2x + 5) − (x² − 5x + 1)(2)] / (2x + 5)²
Expand the numerator:
(2x − 5)(2x + 5) = 4x² − 25
(x² − 5x + 1)(2) = 2x² − 10x + 2
Subtract: (4x² − 25) − (2x² − 10x + 2) = 2x² + 10x − 27
So h'(x) = (2x² + 10x − 27) / (2x + 5)².
Christa wrote the quotient rule as:
d/dx [f(x)/g(x)] = [d/dx f(x)] / [d/dx g(x)]
This is wrong. She just differentiated the top and bottom separately, which is not how the quotient rule works. This is the single most common quotient-rule error.
The quotient rule is not "derivative of the top over derivative of the bottom." It requires the cross-multiplication structure described above. There is no shortcut that lets you differentiate numerator and denominator independently.
This is a quotient where both the numerator and denominator involve named functions. You need the quotient rule, and inside it the sum rule for the denominator.
Let the numerator be g(x) and the denominator be 2x + f(x).
Step 1: Identify the pieces
"Top" = g(x), so d/dx(top) = g'(x)
"Bottom" = 2x + f(x), so d/dx(bottom) = 2 + f'(x)
Step 2: Apply the quotient rule
dh/dx = [g'(x)·(2x + f(x)) − g(x)·(2 + f'(x))] / [2x + f(x)]²
That is the final answer. You leave it in terms of f, g, f', and g' because you are not given explicit formulas for them.
Two functions multiplied together: product rule
One function divided by another: quotient rule
A function inside another function: chain rule (coming soon)
A sum or difference of functions: just differentiate term by term
You are given the graph of f(x) and asked to sketch f'(x). You do this by reading slopes.
Where f is going up (positive slope), plot f' above the x-axis
Where f is going down (negative slope), plot f' below the x-axis
Where f has a local max or min (slope = 0), f' crosses the x-axis
Where f is a straight line, f' is a horizontal line (constant slope)
Where f has a corner, cusp, or vertical tangent, f' does not exist: leave an open circle or gap in the f' graph at that x-value
The steeper the graph of f, the larger the magnitude of f'. A gently rising curve gives small positive values for f'. A steeply falling curve gives large negative values for f'.
The graphs in Problem 3 include piecewise-linear segments and curves. For straight-line segments, the derivative is constant (a flat line at whatever that slope is). At the corners where line segments meet, the derivative jumps, meaning f' has a discontinuity (the derivative does not exist at those junction points).
For smooth curves, estimate the slope at several points: where is it zero, where is it most positive, where is it most negative? Sketch a smooth curve through those estimates.
For f(x) = xᵃ, the derivative is f'(x) = a·xᵃ⁻¹. The value of a controls everything about the derivative's behaviour.
f'(−1) = a·(−1)ᵃ⁻¹
For this to be positive, you need a·(−1)ᵃ⁻¹ > 0. If a is an even positive integer, then (−1)ᵃ⁻¹ = −1, so a·(−1) < 0. That does not work.
If a is an odd positive integer (say a = 3), then (−1)² = 1, so f'(−1) = 3·1 = 3 > 0. Any odd positive integer works.
More generally, a can be any even integer (including 2, 4, 6, ...), since then a − 1 is odd, making (−1)ᵃ⁻¹ = −1, and... wait. Let us be precise: a = 2 gives f'(−1) = 2·(−1)¹ = −2 < 0. a = 3 gives f'(−1) = 3·(−1)² = 3 > 0. So a = 3 (or any odd positive integer) works.
We need two conditions:
f'(−1) = a·(−1)ᵃ⁻¹ < 0
f'(0) = a·0ᵃ⁻¹ does not exist, which requires a − 1 < 0, i.e. a < 1
If a is a negative odd integer, say a = −1, then f'(x) = −x⁻², so f'(−1) = −1 < 0 and f'(0) does not exist (division by zero). This works.
The key insight: f'(0) fails to exist when a < 1, because the exponent a − 1 becomes negative, creating a term like 1/x^(something) that blows up at zero.
Students often apply a fake "quotient rule" by differentiating the numerator and denominator separately: d/dx [f/g] ≠ f'/g'. This was exactly Christa's mistake in Problem 8. The correct formula requires the cross terms.
Forgetting the minus sign in the quotient rule. The numerator is f'g − fg', not f'g + fg'. The subtraction matters.
Forgetting to square the denominator. The quotient rule denominator is [g(x)]², not g(x).
When graphing derivatives, students sometimes plot the y-values of f instead of the slopes. The derivative graph shows slopes, not heights. If f passes through (3, 5) with a slope of 2, you plot the point (3, 2) on the f' graph, not (3, 5).
Assuming f'(x) does not exist only at discontinuities. The derivative also fails to exist at corners, cusps, and vertical tangent lines, even when f itself is continuous.
⚠️ "Identify the mistake" problems like Problem 8 (Christa's incorrect quotient rule) are a recurring exam format. You may be given someone's work and asked to find the error. Know the quotient rule's exact structure.
⚠️ Graphing the derivative from a graph of f is a very common exam question that requires no computation, only interpretation. Practise reading slopes off graphs until it becomes automatic.
⚠️ Problems that give you named functions (f, g, h) without explicit formulas and ask you to differentiate a combination using the product/quotient rules test whether you understand the rules abstractly, not just mechanically. Your answer will be in terms of f, g, f', g'.
⚠️ Power function problems ("find a value of a such that...") test conceptual understanding of how the exponent controls derivative behaviour. Pay attention to what happens at x = 0 and at negative x-values.
True or False: d/dx [f(x)/g(x)] = f'(x)/g'(x).
Answer: False. That is the most common quotient-rule mistake.
Fill in the blank: When the graph of f has a local maximum, the graph of f' crosses the ______.
Answer: x-axis (f' = 0 at a local max).
True or False: If f(x) = x^(1/3), then f'(0) exists.
Answer: False. f'(x) = (1/3)x^(−2/3), which is undefined at x = 0.
Fill in the blank: The denominator of the quotient rule is always [g(x)] raised to the power ______.
Answer: 2 (squared).
Q: State the quotient rule for d/dx [f(x)/g(x)].
A: [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]².
Q: Christa claims that d/dx [(x² − 5x + 1)/(2x + 5)] = (2x − 5)/2. What is her mistake, and what is the correct derivative?
A: She differentiated the numerator and denominator separately, which is not the quotient rule. The correct derivative is (2x² + 10x − 27) / (2x + 5)².
Q: From the graph of a function f, you observe that f is decreasing and getting steeper as x increases. What does this tell you about f'(x)?
A: f'(x) is negative (because f is decreasing) and becoming more negative (because the slope is getting steeper in the negative direction).
Q: For f(x) = xᵃ, find a value of a such that f'(−1) > 0.
A: a = 3 works, since f'(x) = 3x², so f'(−1) = 3 > 0. Any odd positive integer works.
Q: For f(x) = xᵃ, find a value of a such that f'(−1) < 0 and f'(0) does not exist.
A: a = −1 works, since f'(x) = −x⁻², so f'(−1) = −1 < 0 and f'(0) is undefined.
The quotient rule becomes essential when you hit related rates and implicit differentiation later in the course. Many real-world rate-of-change problems involve ratios of changing quantities, so the quotient rule will keep turning up.
Graphing derivatives connects directly to curve sketching (Chapter 4), where you use the sign of f' to determine intervals of increase/decrease and the sign of f'' for concavity. The skill of reading slopes off a graph is the foundation for all of that analysis.
Power functions with fractional exponents connect to the chain rule (coming soon), because expressions like √(x² + 1) require differentiating an outer power applied to an inner function.
quotient rule, product rule, differentiation rules, graphing derivatives, derivative from a graph, slope of tangent line, differentiability, corners, cusps, vertical tangent, power function, f(x) = x^a, horizontal tangent, MTH 132, Calculus I, MSU, Michigan State, Recitation 4, Sections 2.2 and 2.3