AP Calculus AB Differentiation: Definition and Fundamental Properties — Worked Answer Explanations

Unit 2 · 12% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Calculus AB Differentiation: Definition and Fundamental Properties practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Differentiation: Definition and Fundamental Properties practice test and come back here to review, or head back to the Differentiation: Definition and Fundamental Properties unit overview.

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Question 1 · Easy
Find the derivative of $f (x) = 5 x^{3} - 4 x^{2} + 7 x - 2$ .
- A
  $15 x^{2} - 8 x + 7$ Correct
- B
  $15 x^{3} - 8 x^{2} + 7$
  Why not B: Multiplied coefficients by exponents but kept exponents the same.
- C
  $5 x^{2} - 4 x + 7$
  Why not C: Lowered exponents but didn't multiply by them.
- D
  $15 x^{2} - 8 x + 7 - 2 x$
  Why not D: Wrongly differentiated the constant term.
Explanation
Power rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$ . Term-by-term: $15 x^{2} - 8 x + 7 + 0 = 15 x^{2} - 8 x + 7$ .

Key takeaway
Derivative of a constant is zero; power rule applies to each polynomial term.
Question 2 · Easy
If $f (x) = x sin x$ , find $f^{'} (x)$ .
- A
  $sin x + x cos x$ Correct
- B
  $cos x$
  Why not B: Forgot the product rule entirely.
- C
  $x cos x$
  Why not C: Forgot the first term in the product rule.
- D
  $sin x cos x$
  Why not D: Combined two functions incorrectly.
Explanation
Product rule: $(uv)^{'} = u^{'} v + u v^{'}$ . Here $u = x$ , $v = sin x$ , $u^{'} = 1$ , $v^{'} = cos x$ . So $f^{'} = sin x + x cos x$ .

Key takeaway
Product rule: $(fg)' = f'g + fg'$.
Question 3 · Easy
If the position of a particle is $s (t) = t^{3} - 6 t^{2} + 9 t$ (in meters, $t$ in seconds), what is its velocity at $t = 2$ ?
- A
  $- 3 m/s$ Correct
- B
  $- 1 m/s$
  Why not B: Computed $s (2)$ instead of $s^{'} (2)$ .
- C
  $2 m/s$
  Why not C: Plugged $t$ into wrong derivative.
- D
  $9 m/s$
  Why not D: Used initial velocity, $s^{'} (0) = 9$ .
Explanation
$v (t) = s^{'} (t) = 3 t^{2} - 12 t + 9$ . $v (2) = 12 - 24 + 9 = - 3 m/s$ .

Key takeaway
Velocity is the derivative of position with respect to time.
Question 4 · Easy
Find the equation of the tangent line to $f (x) = x^{2}$ at $x = 3$ .
- A
  $y = 6 x - 9$ Correct
- B
  $y = 6 x + 9$
  Why not B: Sign error in y-intercept.
- C
  $y = 9 x - 18$
  Why not C: Used $f (3)$ as the slope.
- D
  $y = 3 x - 9$
  Why not D: Used $x$ as slope.
Explanation
$f^{'} (x) = 2 x$ , so $f^{'} (3) = 6$ (slope). $f (3) = 9$ (point). Equation: $y - 9 = 6 (x - 3) \Rightarrow y = 6 x - 9$ .

Key takeaway
Tangent line: $y - f(a) = f'(a)(x - a)$.
Question 5 · Easy
If $f$ is differentiable at $x = a$ , then $f$ is:
- A
  Continuous at $x = a$ .Correct
- B
  Equal to its derivative at $x = a$ .
  Why not B: $f (a)$ and $f^{'} (a)$ are unrelated quantities.
- C
  Continuous at every $x$ .
  Why not C: Differentiability at one point doesn't imply continuity everywhere.
- D
  Linear near $x = a$ .
  Why not D: Tangent is linear, but $f$ itself need not be.
Explanation
Differentiability implies continuity. The converse fails (e.g., $∣ x ∣$ is continuous at 0 but not differentiable).

Key takeaway
Differentiable ⇒ continuous, but continuous does not ⇒ differentiable.
Question 6 · Easy
Find $f^{'} (x)$ if $f (x) = e^{x} cos x$ .
- A
  $e^{x} (cos x - sin x)$ Correct
- B
  $e^{x} cos x$
  Why not B: Forgot product rule.
- C
  $- e^{x} sin x$
  Why not C: Forgot first term of product rule.
- D
  $e^{x} (cos x + sin x)$
  Why not D: Sign error on $sin x$ derivative.
Explanation
Product rule: $(e^{x})^{'} (cos x) + (e^{x}) (cos x)^{'} = e^{x} cos x + e^{x} (- sin x) = e^{x} (cos x - sin x)$ .

Key takeaway
Remember $(\cos x)' = -\sin x$ — sign matters.
Question 7 · Medium
Find $\frac{d}{d x} [\frac{x ^{2}}{x + 1}]$ .
- A
  $2 x$
  Why not A: Treated as just $x^{2}$ .
- B
  $\frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}$ Correct
- C
  $\frac{2 x}{x + 1}$
  Why not C: Forgot the second term in the quotient rule.
- D
  $\frac{2 x + 1}{x + 1}$
  Why not D: Confused with adding instead of differentiating quotient.
Explanation
Quotient rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ . Here $u = x^{2}, u^{'} = 2 x, v = x + 1, v^{'} = 1$ . Numerator: $2 x (x + 1) - x^{2} (1) = 2 x^{2} + 2 x - x^{2} = x^{2} + 2 x$ .

Key takeaway
Quotient rule: $(u/v)' = (u'v - uv')/v^2$. Always square the bottom.
Question 8 · Medium
Use the limit definition of the derivative to find $f^{'} (2)$ for $f (x) = x^{2} + 1$ .
- A
  $0$
  Why not A: Treated $f (x)$ as a constant.
- B
  $2$
  Why not B: Used $f (2) = 5$ , then divided.
- C
  $4$ Correct
- D
  $5$
  Why not D: Reported $f (2)$ instead of $f^{'} (2)$ .
Explanation
$f^{'} (2) = lim_{h \to 0} \frac{( 2 + h ) ^{2} + 1 - 5}{h} = lim_{h \to 0} \frac{4 h + h ^{2}}{h} = lim_{h \to 0} (4 + h) = 4$ .

Key takeaway
Definition: $f'(a) = \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$.