Learning Objectives
Select derivative rules from an expression's structure.
Combine differentiation with tangent-line and domain reasoning.
Explain derivative existence and interpret an instantaneous rate.
Prerequisites: the five derivative lessons in this section. Trigonometric arguments are in radians. State where derivative formulas are valid, especially when denominators, roots, or implicit relations are involved.
1. Identify the Outermost Operation
Structure | First rule |
|---|---|
Sum of functions | Differentiate term by term |
Product of changing factors | Product rule |
Quotient | Simplify or use quotient rule |
One function inside another | Chain rule |
A relation involving x and y | Differentiate both sides implicitly |
A question about derivative existence | Return to one-sided difference quotients and continuity |
A complicated expression often uses several rules. Begin with the outermost structure, then handle each component. For x² sin(3x), start with the product rule and use the chain rule only when differentiating sin(3x).
2. Keep the Point and Slope Separate
Worked Example: A Tangent to a Composition
Find the tangent line to f(x) = (x² + 1)³ at x = 1. The function value is f(1) = 8.
The chain rule gives f′(x) = 3(x² + 1)²·2x = 6x(x² + 1)². At x = 1, the slope is 24.
Therefore y − 8 = 24(x − 1), or y = 24x − 16. The derivative value 24 is the slope, not the y-coordinate of the point.
3. Use Mathematical Checks
A derivative of a polynomial should also be a polynomial of one lower degree, unless the original is constant. For a differentiable even function, its derivative is odd wherever the symmetry applies. Alternative algebraic forms can provide checks, but numerical agreement at one input alone does not prove a general derivative formula.
Common Mistakes
A continuous function may have a corner and fail to be differentiable.
The derivative of a product is not the product of derivatives.
A missing chain-rule factor changes the answer even when the outer derivative is correct.
In an implicit relation, y-dependent terms must include their dependence on x.
4. Mixed Practice
1. Differentiate f(x) = 4x⁵ − 3/x + 2.
2. Differentiate g(x) = x² cos x.
3. Differentiate h(x) = (x + 1)/(x − 1).
4. Differentiate p(x) = e^(x² − 3x).
5. Find the tangent line to y = ln x at x = 1.
6. For x² + xy = 6, find dy/dx and its value at (2, 1).
7. Let f(x) = x² for x < 1 and f(x) = 2x − 1 for x ≥ 1. Is f differentiable at 1? Explain.
8. If C(q) is a cost in dollars for q kilograms produced, interpret C′(10) = 7.
Worked Solutions
1. Rewrite −3/x as −3x⁻¹. The derivative is f′(x) = 20x⁴ + 3/x², with x ≠ 0.
2. The product rule gives g′(x) = 2x cos x − x² sin x.
3. h′(x) = [(x − 1) − (x + 1)]/(x − 1)² = −2/(x − 1)², with x ≠ 1.
4. Apply the chain rule: p′(x) = (2x − 3)e^(x² − 3x).
5. The point is (1, 0) and the slope is 1/1 = 1. The tangent line is y = x − 1.
6. Differentiate to get 2x + y + xy′ = 0. Thus y′ = −(2x + y)/x for x ≠ 0. The point satisfies 4 + 2 = 6, and its slope is −5/2.
7. Both pieces meet at value 1, so the function is continuous. For h < 0, the difference quotient at 1 is [(1 + h)² − 1]/h = 2 + h, approaching 2. For h > 0, it is [2(1 + h) − 1 − 1]/h = 2. Both sides agree, so f′(1) = 2.
8. At an output level of 10 kg, cost has an instantaneous rate of increase of $7 per additional kilogram. This is a local rate, not a claim that total cost is $7 or that every future kilogram costs exactly $7.
5. Summary
- Choose derivative rules from the outer structure, then work inward.
- Keep domain restrictions and chain-rule factors.
- A tangent line combines a function value with a derivative value.
- Use definitions and units to justify existence and interpretation.