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Grade 12 · unit review

Review: 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.

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Unit review questions
1. For f(x)=x², find f′(7) from the limit of [f(7+h)−f(7)]/h.
Need a hint?
  • A difference quotient measures the secant slope over a nonzero input change.
  • Do not substitute zero into the original quotient before simplifying, because it divides by the change.
2. For f(x)=x³, find f′(7).
Need a hint?
  • For a power x^n, the derivative is n x^(n−1) where the rule applies.
  • Differentiate before substituting the evaluation input.
3. For f(x)=x²(x+1), find f′(7).
Need a hint?
  • The derivative of uv is u′v+uv′.
  • Multiplying u′ and v′ is not the product rule.
4. For y=x², find the tangent line at x=7.
Need a hint?
  • A tangent line at an input passes through that function point and has derivative slope there.
  • A correct slope alone does not locate the tangent line.

The download contains your current choices and any checked feedback. Keep the separate transfer-task work in your own document; downloading its instructions does not mark it complete.

Use the skills in a new task

Create a four-part mathematical solution sheet for these supplied cases. 1. For f(x)=x², find f′(9) from the limit of [f(9+h)−f(9)]/h. Candidate answers: (A) 17 (B) 18 (C) 19 (D) 36. 2. For f(x)=x³, find f′(9). Candidate answers: (A) 242 (B) 243 (C) 244 (D) 486. 3. For f(x)=x²(x+1), find f′(9). Candidate answers: (A) 260 (B) 261 (C) 262 (D) 522. 4. For y=x², find the tangent line at x=9. Candidate answers: (A) y+81=18(x+9) (B) y=18x (C) y−81=18(x−9) (D) y−81=9(x−9).

Work on paper or in your own document. Explain the choices you made.

  • State each requested quantity or relationship and its input restrictions.
  • Show intermediate calculations or the applicable theorem with matching correspondences.
  • Give all four results, including exact values or requested rounding and units.
  • Check a result by substitution, reversal, dimensions, a representation or a counterexample as appropriate.
Compare with a model response

1. 18. Expand: [(9+h)²−81]/h=[18h+h²]/h=18+h for h≠0. Taking h→0 gives 18. The derivative is the limiting secant slope. 2. 243. The power rule gives f′(x)=3x². Substitution x=9 yields 3(9)²=243. The exponent decreases by one; it is not divided by three. 3. 261. The product rule gives f′(x)=2x(x+1)+x²=3x²+2x. At x=9, this is 261. Differentiating each factor and multiplying would omit essential terms. 4. y−81=18(x−9). The point is (9,81) and derivative slope is 2(9)=18. Insert both into point-slope form. The tangent is a local linear model, not the whole parabola.

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