Difficulty: Intermediate | Prerequisites: Limits (see Limits study notes), algebra, basic trigonometry
The derivative measures the instantaneous rate of change of a function, defined as a limit of the difference quotient. Differentiation techniques (product rule, quotient rule, chain rule, implicit differentiation) let you find derivatives efficiently, and once you have them, you can locate extrema and inflection points, which is where most exam problems land.
Derivative
The instantaneous rate of change of f(x) at a point x, defined as lim (h→0) [f(x+h) − f(x)] / h. Notations include f'(x), dy/dx, and Df.
In simple terms, this is the slope of the curve at a single point.
Difference quotient
[f(x+h) − f(x)] / h. The expression whose limit as h → 0 gives the derivative.
Think of it as the slope of a secant line that gets shorter and shorter until it becomes a tangent line.
Product rule
If y = u·v, then y' = u'v + uv'.
In simple terms: differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
Quotient rule
If y = u/v, then y' = (u'v − uv') / v².
Think of it as "low d-high minus high d-low, over the square of what's below."
Chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). Differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
In simple terms: peel the layers one at a time, from outside in.
Implicit differentiation
A technique for finding dy/dx when y is not isolated on one side of the equation. Differentiate both sides with respect to x, treating y as a function of x (so every term with y picks up a dy/dx by the chain rule), then solve for dy/dx.
Critical point
A value of x where f'(x) = 0 or f'(x) is undefined. Candidates for local maxima and minima.
Inflection point
A point where the concavity of f changes, i.e. where f''(x) changes sign.
Think of it as where the curve switches from bending one way to bending the other.
Local minimum / local maximum
A point where f(x) is smaller (or larger) than at all nearby points. Found using the first or second derivative test.
Absolute minimum / absolute maximum
The smallest (or largest) value of f(x) on a closed interval. Found by evaluating f at all critical points in the interval and at both endpoints, then comparing.
Worked example (Problem 7): Let f(x) = 4x − 7x². Find f'(x) using the definition.
Compute f(x+h) = 4(x+h) − 7(x+h)² = 4x + 4h − 7x² − 14xh − 7h².
f(x+h) − f(x) = 4h − 14xh − 7h².
Divide by h: 4 − 14x − 7h.
Take the limit as h → 0: f'(x) = 4 − 14x.
Worked example (Problem 8): Let y = x²e^(−x). Find the local minimum value.
Apply the product rule: y' = 2x·e^(−x) + x²·(−e^(−x)) = xe^(−x)(2 − x).
Set y' = 0. Since e^(−x) is never zero, the critical points are x = 0 and x = 2.
Sign analysis of y': negative for x < 0, positive for 0 < x < 2, negative for x > 2.
x = 0 is a local minimum (y' changes from − to +). The local minimum value is y(0) = 0.
x = 2 is a local maximum. The local maximum value is y(2) = 4/e².
Worked example (Problem 9): y = xy² − 2x². Find the slope of the tangent line at (1, −1).
Differentiate both sides with respect to x: y' = 1·y² + x·2y·y' − 4x.
Rearrange: y' − 2xy·y' = y² − 4x, so y'(1 − 2xy) = y² − 4x.
Solve: y' = (y² − 4x) / (1 − 2xy).
At (1, −1): y' = (1 − 4) / (1 − 2(1)(−1)) = −3 / 3 = −1.
Worked example (Problem 10): Let f(x) = (x² − 8x) / (x + 1). Find the absolute minimum on [0, 4].
Apply the quotient rule: f'(x) = [(2x − 8)(x + 1) − (x² − 8x)(1)] / (x + 1)².
Simplify the numerator: 2x² − 6x − 8 − x² + 8x = x² + 2x − 8 = (x + 4)(x − 2).
f'(x) = 0 at x = −4 (outside [0, 4]) and x = 2.
Evaluate: f(0) = 0, f(2) = (4 − 16)/3 = −4, f(4) = (16 − 32)/5 = −3.2.
The absolute minimum value is −4, occurring at x = 2.
Worked example (Problem 11): Let f(x) = x² − 8x^(3/2). Find the x-value of the inflection point.
f'(x) = 2x − 12x^(1/2).
f''(x) = 2 − 6/√x.
Set f''(x) = 0: 2 = 6/√x, so √x = 3, giving x = 9.
Check: f'' is undefined at x = 0 and changes sign at x = 9 (negative for 0 < x < 9, positive for x > 9), confirming an inflection point.
The inflection point is (9, −135), since f(9) = 81 − 8·27 = −135.
Worked example (Problem 12): Let y = arcsin(√x). Find y'(1/2).
By the chain rule: y' = [1 / √(1 − (√x)²)] · (1 / 2√x) = 1 / [2√x · √(1 − x)].
At x = 1/2: y'(1/2) = 1 / [2 · √(1/2) · √(1/2)] = 1 / [2 · 1/2] = 1.
Definition of the derivative:
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}Power rule: d/dx [xⁿ] = nxⁿ⁻¹
Product rule: d/dx [uv] = u'v + uv'
Quotient rule: d/dx [u/v] = (u'v − uv') / v²
Chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
Derivative of arcsin: d/dx [arcsin(u)] = u' / √(1 − u²)
Derivative of eˣ: d/dx [eˣ] = eˣ
Derivative of ln(x): d/dx [ln(x)] = 1/x
Students often confuse the quotient rule with L'Hôpital's Rule. The quotient rule is for differentiating a ratio of functions. L'Hôpital's Rule is for evaluating limits of indeterminate forms. They are completely different operations.
Students often forget to apply the chain rule when differentiating composite functions. If there is a "function inside a function," you must multiply by the derivative of the inner function.
Students often assume that a critical point is automatically a maximum or minimum. It is only a candidate. You must check with the first or second derivative test.
Students often set f'(x) = 0 and forget to check where f'(x) is undefined. Both types of points are critical points.
⚠️ "Find the derivative using the definition" is a standard exam question. You must show the full limit process, not just apply the power rule.
⚠️ Implicit differentiation problems appear frequently. Remember: every time you differentiate a y term, you attach a dy/dx.
⚠️ Closed-interval absolute extremum problems follow a fixed procedure: find critical points, evaluate f at each critical point and at both endpoints, compare. Do not skip the endpoints.
⚠️ Inflection points require f'' to change sign at the candidate. Simply having f''(x) = 0 is necessary but not sufficient.
True or false: The derivative of f(x) = 4x − 7x² is 4 − 7x.
False. The correct derivative is 4 − 14x (apply the power rule to −7x²).
Fill in the blank: To find the absolute minimum of f on [a, b], evaluate f at all ______ in (a, b) and at both ______.
Critical points; endpoints.
True or false: If f''(c) = 0, then x = c is an inflection point.
False. f''(c) = 0 is necessary but f'' must also change sign at c.
Fill in the blank: When differentiating y = arcsin(u), the result is ______.
u' / √(1 − u²).
True or false: In implicit differentiation, every differentiated y-term picks up a factor of dy/dx.
True.
Q: Let f(x) = 4x − 7x². Find f'(x) from the definition of the derivative.
A: f'(x) = lim (h→0) [(4(x+h) − 7(x+h)²) − (4x − 7x²)] / h. Expanding and simplifying gives lim (h→0) (4 − 14x − 7h) = 4 − 14x.
Q: Let y = x²e^(−x). What is the local minimum value?
A: y' = xe^(−x)(2 − x). Critical points at x = 0 and x = 2. The local minimum value is y(0) = 0.
Q: Given y = xy² − 2x², find dy/dx at (1, −1).
A: Implicit differentiation gives y' = (y² − 4x) / (1 − 2xy). At (1, −1): y' = −3/3 = −1.
Q: Let f(x) = (x² − 8x)/(x + 1). What is the absolute minimum of f on [0, 4]?
A: f'(x) = (x + 4)(x − 2)/(x + 1)². Critical point in [0, 4] is x = 2. f(0) = 0, f(2) = −4, f(4) = −3.2. Absolute minimum is −4.
Q: Let f(x) = x² − 8x^(3/2). At what x-value does the inflection point occur?
A: f''(x) = 2 − 6/√x = 0 gives √x = 3, so x = 9.
Derivatives are built on limits (see the Limits study notes). The definition of the derivative is itself a limit, and L'Hôpital's Rule uses derivatives to evaluate limits, so the two topics feed into each other.
Derivatives connect forward to integrals through the Fundamental Theorem of Calculus: integration reverses differentiation. The antiderivative techniques in the Integrals study notes are essentially differentiation run backwards.
Optimisation problems (finding maxima and minima) appear across physics, engineering, and economics. The critical-point and closed-interval methods covered here are the tools for all of them.
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