Difficulty: Beginner | Prerequisites: Power rule for derivatives, exponent algebra
Tags: sum rule, difference rule, constant multiple rule, tangent line, point-slope form, calculus I, brief calculus, differentiation, polynomial derivative, ASU
The power rule tells you how to differentiate a single term like xⁿ. Most functions you actually meet are built from several such terms added, subtracted, or scaled by constants. The sum, difference, and constant multiple rules let you break any polynomial (or similar expression) into its individual terms, differentiate each one, and reassemble.
Once you can differentiate a function, the immediate payoff is finding the equation of a tangent line at any point on its curve. Tangent-line problems tie together everything from this unit and appear on nearly every exam.
You should already be comfortable with the power rule, including negative and fractional exponents.
You can differentiate term by term: the derivative of a sum is the sum of the derivatives, and constants pull straight out. To find a tangent line, evaluate the derivative at the point to get the slope, then plug into point-slope form.
Sum rule
The derivative of f(x) + g(x) equals f'(x) + g'(x). In simple terms, differentiate each piece separately and add the results.
Difference rule
The derivative of f(x) − g(x) equals f'(x) − g'(x). Same idea as the sum rule, just with a minus sign.
Constant multiple rule
The derivative of c · f(x) equals c · f'(x), where c is any constant. Think of it as: constants pass through differentiation untouched.
Tangent line
A straight line that just touches a curve at a single point and has the same slope as the curve at that point. The derivative gives you that slope.
Point-slope form
The equation y = f'(a)(x − a) + f(a), used to write the tangent line at x = a. You need two things: the slope f'(a) and the y-value f(a).
[f(x) ± g(x)]' = f'(x) ± g'(x)
Differentiate each term independently, then combine.
[c · f(x)]' = c · f'(x)
Pull the constant out, differentiate what remains.
Example 1: f(x) = 5x⁷ − 2x³ + 1
Three separate terms. Differentiate each:
5 · 7x⁶ = 35x⁶
−2 · 3x² = −6x²
The constant 1 gives 0
Result: f'(x) = 35x⁶ − 6x²
Example 2: g(x) = 1/x⁴ − 7/x⁵
Rewrite: g(x) = x⁻⁴ − 7x⁻⁵
Differentiate:
−4x⁻⁵
−7 · (−5)x⁻⁶ = 35x⁻⁶
Result: g'(x) = −4x⁻⁵ + 35x⁻⁶ = −4/x⁵ + 35/x⁶
Example 3: h(x) = √x − 1/³√x + ³√9
Rewrite: h(x) = x^(1/2) − x^(−1/3) + ³√9
The term ³√9 is a constant (no variable), so its derivative is 0.
Differentiate:
(1/2)x^(−1/2)
−(−1/3)x^(−4/3) = (1/3)x^(−4/3)
Result: h'(x) = 1/(2√x) + 1/(3·³√x⁴)
The tangent line at x = a has two ingredients: the slope (which is f'(a)) and a point on the curve ((a, f(a))). Plug both into point-slope form.
Steps:
Find f'(x) using the differentiation rules.
Evaluate f'(a) to get the slope.
Evaluate f(a) to get the y-coordinate.
Write the tangent line: y = f'(a)(x − a) + f(a).
Example 1: f(t) = 1/(3t) at the point (2, 1/6)
Rewrite: f(t) = (1/3)t⁻¹
Differentiate: f'(t) = −(1/3)t⁻² = −1/(3t²)
Slope at t = 2: f'(2) = −1/(3 · 4) = −1/12
Point-slope: y = (−1/12)(t − 2) + 1/6
Simplify: y = (−1/12)t + 1/3
Example 2: f(x) = x² + √x + 1 at x = 9
Differentiate: f'(x) = 2x + (1/2)x^(−1/2)
Slope at x = 9: f'(9) = 18 + (1/2)(1/3) = 18 + 1/6 = 109/6
y-value: f(9) = 81 + 3 + 1 = 85
Tangent line: y = (109/6)(x − 9) + 85
[f(x) \pm g(x)]' = f'(x) \pm g'(x)[c \cdot f(x)]' = c \cdot f'(x)\text{Tangent line at } x = a: \quad y = f'(a)(x - a) + f(a)Tangent lines are used constantly in linear approximation, where you estimate a function's value near a known point by using the tangent line instead of the full function. Economists use this when approximating marginal changes in cost or revenue near current production levels.
Students sometimes try to "distribute" differentiation over a product, writing (f · g)' = f' · g'. This is wrong. The sum/difference rules apply to addition and subtraction, not multiplication. Products require the product rule (a later topic).
When differentiating a sum, students occasionally change the signs between terms. The sign in front of each term carries through exactly as it appears.
In tangent-line problems, students often forget to compute f(a) for the y-coordinate and leave it out of the equation. You need both the slope and the point.
A frequent error is plugging the x-value into the original function to get the slope. The slope comes from f'(a), not f(a).
⚠️ Tangent-line problems are a staple. Expect at least one on any derivatives exam. The method is always the same: find f'(x), evaluate at the point, write point-slope form.
⚠️ "Simplify before differentiating" problems test whether you can rewrite a quotient or product into individual power terms before applying the rules. Do not skip this step.
⚠️ Watch for terms that look complicated but are actually constants (like ³√9). Their derivative is zero.
True or false: The derivative of f(x) + g(x) is f'(x) · g'(x).
Fill in the blank: If f(x) = 3x⁸, then f'(x) = ____.
True or false: To find a tangent line you need both the slope and a point on the curve.
Fill in the blank: The slope of the tangent line at x = a is given by ____.
True or false: The derivative of 7f(x) is 7f'(x).
Answers: 1. False (it is f'(x) + g'(x)). 2. 24x⁷. 3. True. 4. f'(a). 5. True.
Q: Differentiate f(x) = 5x⁷ − 2x³ + 1.
A: f'(x) = 35x⁶ − 6x². The constant 1 drops away.
Q: Find the derivative of g(x) = 1/x⁴ − 7/x⁵. Write with positive exponents.
A: Rewrite as x⁻⁴ − 7x⁻⁵. Then g'(x) = −4x⁻⁵ + 35x⁻⁶ = −4/x⁵ + 35/x⁶.
Q: Find the equation of the tangent line to f(t) = 1/(3t) at the point (2, 1/6).
A: f'(t) = −1/(3t²). At t = 2, slope = −1/12. Tangent line: y = (−1/12)(t − 2) + 1/6, which simplifies to y = (−1/12)t + 1/3.
Q: Find the tangent line to f(x) = x² + √x + 1 at x = 9.
A: f'(x) = 2x + (1/2)x^(−1/2). At x = 9, slope = 18 + 1/6 = 109/6. Since f(9) = 85, the tangent line is y = (109/6)(x − 9) + 85.
Q: Differentiate h(x) = (8x⁴ + 18x²) / (2x⁴).
A: Simplify first: h(x) = 4 + 9x⁻¹. Then h'(x) = −9x⁻² = −9/x².
These rules build directly on the power rule covered in the companion study notes. Together they let you differentiate any polynomial, and they form the foundation for the product rule, quotient rule, and chain rule that come next.
Tangent lines come back when you study linear approximation, optimisation (finding where the tangent is horizontal), and related rates.
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