Study guide: Introductory Calculus: Derivatives and Local Lines
A derivative measures instantaneous rate through a limiting difference quotient. Differentiation rules summarize that process for suitable functions, while a tangent line uses the derivative slope and the function’s actual point to model local behavior.
Ideas to explain in your own words
- A difference quotient measures the secant slope over a nonzero input change. Expanding and canceling common terms reveals a simpler expression whose limit as the change tends to zero gives the derivative. Do not substitute zero into the original quotient before simplifying, because it divides by the change.
- For a power x^n, the derivative is n x^(n−1) where the rule applies. Multiplying by the exponent and lowering the power produces the cubic’s instantaneous slope at each input. Differentiate before substituting the evaluation input.
- The derivative of uv is u′v+uv′. Each factor’s change contributes while the other factor is retained, which explains why the two terms must be added. Multiplying u′ and v′ is not the product rule.
- A tangent line at an input passes through that function point and has derivative slope there. Point-slope form combines these two pieces into a local linear approximation. A correct slope alone does not locate the tangent line.
Useful vocabulary
- difference quotient
- A finite output change divided by its nonzero input change.
- derivative
- The limiting instantaneous rate of change when that limit exists.
- tangent line
- A local line through a curve point with its derivative slope.
Watch your reasoning
- Do not substitute zero into the original quotient before simplifying, because it divides by the change. Cancel only for nonzero change, then take the limit; the derivative is not just a finite-step average.
- Differentiate before substituting the evaluation input. The power rule changes both coefficient and exponent; simply dividing by the exponent or retaining the original exponent would produce a different expression.
- Multiplying u′ and v′ is not the product rule. Expand first as an independent check when possible, then confirm both approaches give the same derivative polynomial and evaluation value.
A worked review example
Complete supplied guide examples: Derive a Quadratic’s Instantaneous Slope — For f(x)=x², find f′(2) from the limit of [f(2+h)−f(2)]/h. Candidate answers: (A) 3 (B) 4 (C) 5 (D) 8. Expand: [(2+h)²−4]/h=[4h+h²]/h=4+h for h≠0. Taking h→0 gives 4. The derivative is the limiting secant slope. Apply the Power Rule to a Cubic — For f(x)=x³, find f′(2). Candidate answers: (A) 11 (B) 12 (C) 13 (D) 24. The power rule gives f′(x)=3x². Substitution x=2 yields 3(2)²=12. The exponent decreases by one; it is not divided by three. Differentiate a Product with Both Contributions — For f(x)=x²(x+1), find f′(2). Candidate answers: (A) 15 (B) 16 (C) 17 (D) 32. The product rule gives f′(x)=2x(x+1)+x²=3x²+2x. At x=2, this is 16. Differentiating each factor and multiplying would omit essential terms. Combine a Derivative Slope with a Point — For y=x², find the tangent line at x=2. Candidate answers: (A) y+4=4(x+2) (B) y=4x (C) y−4=2(x−2) (D) y−4=4(x−2). The point is (2,4) and derivative slope is 2(2)=4. Insert both into point-slope form. The tangent is a local linear model, not the whole parabola.
Make a practice plan
- Study the four methods in the listed order; attempt each independent check before opening its explanation.
- Complete each lesson’s three changed-input questions and record the operation or reasoning step behind any correction.
- Take the four-question unit review, then solve the transfer task with written justifications.
- After correcting errors, attempt the separate unit exam and revisit the linked lesson for any unresolved method.