Difficulty: Introductory | Prerequisites: Section 2.1 (limits), Section 2.2 (definition of the derivative)
Section 2.3 is where calculus shifts from computing derivatives the hard way (using the limit definition every single time) to using shortcut formulas that handle polynomials, rational functions, and roots in seconds. These rules, collectively called the differentiation formulas, are the toolkit you will rely on for every calculus problem from here forward. If Section 2.2 was learning to start a fire by rubbing sticks, this section hands you a lighter.
You should already be comfortable with the limit definition of the derivative, basic limit evaluation, and the idea that f'(x) gives the slope of the tangent line. If those feel shaky, revisit Sections 2.1 and 2.2 first.
Instead of computing limits every time, you can differentiate most functions using a handful of rules: the power rule, constant rule, sum/difference rule, product rule, and quotient rule. These formulas let you find first, second, and higher-order derivatives quickly, and from there you can write equations of tangent and perpendicular lines.
Derivative (f'(x) or dy/dx)
The instantaneous rate of change of a function at a point. Geometrically, it is the slope of the tangent line to the curve at that point.
In simple terms, it tells you how fast the output is changing at any given input.
Power rule
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹, for any real number n.
Think of it as: bring the exponent down as a multiplier, then reduce the exponent by one.
Constant rule
The derivative of any constant is zero: if f(x) = c, then f'(x) = 0.
In simple terms, a flat line has no slope.
Constant multiple rule
If f(x) = c · g(x), then f'(x) = c · g'(x). Constants just ride along.
Sum/Difference rule
The derivative of a sum (or difference) is the sum (or difference) of the derivatives: (f ± g)' = f' ± g'.
Think of it as: differentiate each term separately, then combine.
Product rule
If y = u · v, then y' = u'v + uv'.
In simple terms, differentiate each factor in turn while leaving the other alone, then add the results.
Quotient rule
If y = u/v, then y' = (u'v − uv') / v².
Think of it as: "low d-high minus high d-low, all over the square of what's below."
Higher-order derivative
The derivative of a derivative. f''(x) is the second derivative, f'''(x) the third, and f⁽ⁿ⁾(x) the nth.
In simple terms, keep differentiating the result to get successive derivatives.
Tangent line
A line that touches a curve at exactly one point (locally) and has the same slope as the curve there. Equation: y − y₁ = m(x − x₁), where m = f'(x₁).
Perpendicular line (normal line)
A line perpendicular to the tangent line at the point of tangency. Its slope is the negative reciprocal of the tangent slope: m_perp = −1/f'(x₁).
If f(x) = c (any constant), then f'(x) = 0 for all x.
All higher derivatives are also zero: f''(x) = 0, f'''(x) = 0, and so on.
Example: f(x) = 12 → f'(x) = 0, f'(5) = 0, f''(x) = 0, f''(−10) = 0.
For any real exponent n: d/dx [xⁿ] = nxⁿ⁻¹.
This works for positive integers, negative integers, and fractions alike.
x⁴ → 4x³
x⁻¹ → −x⁻²
x^(1/3) → (1/3)x^(−2/3)
x^(1/2) = √x → (1/2)x^(−1/2) = 1/(2√x)
d/dx [c · f(x)] = c · f'(x). The constant factors out.
Example: d/dx [3t⁴] = 3 · 4t³ = 12t³.
(f + g)' = f' + g' and (f − g)' = f' − g'.
Differentiate each term separately.
Example: g(t) = 3t⁴ − 6t² + 4 → g'(t) = 12t³ − 12t.
f(x) = 2 + 4x⁻¹ + 5x⁻²
f'(x) = 0 + 4(−1)x⁻² + 5(−2)x⁻³ = −4x⁻² − 10x⁻³
f'(5) = −4/25 − 10/125 = −4/25 − 2/25 = −6/25
f''(x) = 8x⁻³ + 30x⁻⁴
f''(5) = 8/125 + 30/625 = 8/125 + 6/125 = 14/125
h(x) = 3 − 2x^(1/3)
h'(x) = −2 · (1/3)x^(−2/3) = −(2/3)x^(−2/3)
h'(1) = −2/3
f(x) = 5x + 2√x = 5x + 2x^(1/2)
f'(x) = 5 + 2 · (1/2)x^(−1/2) = 5 + x^(−1/2)
f'(4) = 5 + 1/√4 = 5 + 1/2 = 11/2
g(t) = 3t⁴ − 6t² + 4
g'(t) = 12t³ − 12t
g''(t) = 36t² − 12
g'''(t) = 72t
g⁽⁴⁾(t) = 72
g⁽⁵⁾(t) = 0
If y = u(x) · v(x), then y' = u'v + uv'.
Differentiate the first factor, multiply by the second, then add the first times the derivative of the second.
Worked example:
f(x) = (x^(1/5) + 7)(3x^(1/2) + 7)
Let u = x^(1/5) + 7, so u' = (1/5)x^(−4/5)
Let v = 3x^(1/2) + 7, so v' = (3/2)x^(−1/2)
f'(x) = (1/5)x^(−4/5) · (3x^(1/2) + 7) + (x^(1/5) + 7) · (3/2)x^(−1/2)
You do not need to simplify further on this type of problem.
If y = u/v, then y' = (u'v − uv') / v².
The mnemonic: "low d-high minus high d-low, over the square of what's below."
Worked example:
f(x) = 2/(x − 3)
u = 2, u' = 0
v = (x − 3), v' = 1
f'(x) = (0 · (x − 3) − 2 · 1) / (x − 3)² = −2/(x − 3)²
f'(5) = −2/(5 − 3)² = −2/4 = −1/2
For y = f · g · h, differentiate one factor at a time:
y' = f'gh + fg'h + fgh'
Worked example:
y = (x + 1)(x − 5)(x² + 2)
y' = (1)(x − 5)(x² + 2) + (x + 1)(1)(x² + 2) + (x + 1)(x − 5)(2x)
Again, you do not need to simplify.
Once you have f'(x), you can differentiate again to get f''(x), then f'''(x), and so on. Each successive derivative describes how the previous one is changing.
f'(x) = first derivative (velocity, slope of the tangent)
f''(x) = second derivative (acceleration, concavity)
f'''(x) = third derivative (jerk)
f⁽ⁿ⁾(x) = nth derivative
For a polynomial of degree n, the (n+1)th derivative and all beyond are zero. This is because each differentiation reduces the degree by one, and eventually you differentiate a constant.
Worked example:
g(t) = 3t⁴ − 6t² + 4 (degree 4 polynomial)
g'(t) = 12t³ − 12t (degree 3)
g''(t) = 36t² − 12 (degree 2)
g'''(t) = 72t (degree 1)
g⁽⁴⁾(t) = 72 (degree 0, a constant)
g⁽⁵⁾(t) = 0 (zero from here on)
The tangent line to y = f(x) at x = a has slope m = f'(a) and passes through (a, f(a)). Use point-slope form:
y − f(a) = f'(a)(x − a)
Worked example:
h(x) = 3 − 2x^(1/3), find the tangent at (1, 1).
h'(x) = −(2/3)x^(−2/3)
h'(1) = −2/3
Tangent: y − 1 = (−2/3)(x − 1) → y = (−2/3)x + 5/3
Another example:
f(x) = 2/(x − 3), tangent at (5, 1).
f'(x) = −2/(x − 3)²
f'(5) = −2/4 = −1/2
Tangent: y − 1 = (−1/2)(x − 5) → y = (−1/2)x + 7/2
The perpendicular line at the same point has slope m_perp = −1/f'(a) (the negative reciprocal).
Worked example:
f(x) = 5x + 2√x at x = 4.
f(4) = 20 + 2(2) = 24, so the point is (4, 24)
f'(4) = 11/2
Perpendicular slope = −1/(11/2) = −2/11
Perpendicular line: y − 24 = (−2/11)(x − 4) → y = (−2/11)x + 272/11
A horizontal tangent occurs where f'(x) = 0. Set the derivative equal to zero and solve.
Worked example:
f(x) = 2x³ − 3x² − 36x − 3
f'(x) = 6x² − 6x − 36 = 6(x² − x − 6) = 6(x − 3)(x + 2)
f'(x) = 0 when x = 3 or x = −2
When the problem says the line 9x + y = b is tangent to y = ax³ at x = 1, two conditions must hold simultaneously:
The curve passes through the point: a(1)³ = a, and the line gives y = b − 9, so a = b − 9.
The slopes match: y' = 3ax², so at x = 1 the slope is 3a. The line y = −9x + b has slope −9, so 3a = −9, giving a = −3.
Then b = a + 9 = −3 + 9 = 6.
Rule | Formula |
|---|---|
Constant | d/dx [c] = 0 |
Power | d/dx [xⁿ] = nxⁿ⁻¹ |
Constant multiple | d/dx [cf] = c · f' |
Sum/Difference | d/dx [f ± g] = f' ± g' |
Product | d/dx [uv] = u'v + uv' |
Quotient | d/dx [u/v] = (u'v − uv') / v² |
Tangent line at (a, f(a)) | y − f(a) = f'(a)(x − a) |
Perpendicular slope | m_perp = −1 / f'(a) |
Horizontal tangent | Solve f'(x) = 0 |
The derivative gives the instantaneous rate of change, which is exactly what engineers and scientists need when analysing motion, growth, or decay. Velocity is the derivative of position, acceleration is the derivative of velocity, and marginal cost in economics is the derivative of the total cost function. Every time you hear "rate of change" in any applied field, a derivative is doing the work.
Students often forget that the derivative of a constant is zero, not one. A constant has no rate of change.
When using the power rule on negative or fractional exponents, students frequently forget to subtract 1 from the exponent. For x⁻¹, the derivative is −x⁻², not −x⁰.
The product rule is not "derivative of the first times derivative of the second." You must keep one factor undifferentiated in each term: u'v + uv', not u'v'.
In the quotient rule, the order of subtraction matters. It is u'v minus uv' in the numerator, not the other way round. Swapping them flips the sign of the entire derivative.
⚠️ The power rule with fractional and negative exponents is heavily tested. Be comfortable rewriting √x as x^(1/2) and 1/x² as x⁻² before differentiating.
⚠️ Product rule and quotient rule problems are standard exam fare. Know both formulas cold.
⚠️ Tangent line questions appear on nearly every exam. The process is always the same: compute f'(a) for the slope, find the point, use point-slope form.
⚠️ "Find where the tangent is horizontal" = set f'(x) = 0 and solve. This is a common free-response question.
⚠️ Perpendicular line problems test whether you remember to take the negative reciprocal, not just the derivative.
True or false: The derivative of 7 is 7. _____ (False. It is 0.)
The derivative of x⁵ is _____. (5x⁴)
True or false: d/dx [fg] = f'g'. _____ (False. The product rule gives f'g + fg'.)
The slope of the tangent line to y = x² at x = 3 is _____. (6)
If f'(a) = 4, the slope of the perpendicular line at x = a is _____. (−1/4)
Q: Let f(x) = 4x³ − 2x + 9. Find f'(x).
A: f'(x) = 12x² − 2.
Q: Find the derivative of y = 3/x⁴. (Rewrite as 3x⁻⁴ first.)
A: y' = 3(−4)x⁻⁵ = −12x⁻⁵ = −12/x⁵.
Q: Use the product rule to differentiate y = (2x + 1)(x² − 3).
A: y' = (2)(x² − 3) + (2x + 1)(2x) = 2x² − 6 + 4x² + 2x = 6x² + 2x − 6.
Q: Use the quotient rule to differentiate y = x/(x + 1).
A: y' = ((1)(x + 1) − x(1)) / (x + 1)² = 1/(x + 1)².
Q: Find the equation of the tangent line to y = x² − 3x at x = 2.
A: f(2) = 4 − 6 = −2. f'(x) = 2x − 3, so f'(2) = 1. Tangent: y − (−2) = 1(x − 2), so y = x − 4.
Q: For what values of x does f(x) = x³ − 12x have a horizontal tangent?
A: f'(x) = 3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2). Set f'(x) = 0: x = 2 or x = −2.
Q: If u(0) = 3, u'(0) = −6, v(0) = −4, v'(0) = 7, find d/dx[uv] at x = 0.
A: By the product rule: u'(0)v(0) + u(0)v'(0) = (−6)(−4) + (3)(7) = 24 + 21 = 45.
The differentiation formulas here are the foundation for everything in Chapters 3 and 4. The chain rule (Section 2.5) extends these to composite functions. Related rates (Chapter 3) and optimisation problems (Chapter 4) both depend on being able to differentiate quickly and accurately. If you are slow with these rules, every later topic will feel harder than it needs to be.
The product and quotient rules also reappear in implicit differentiation (Section 2.6), where you differentiate both sides of an equation with respect to x.
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