AP Calculus BC 2.8 The Product Rule - MCQs - Exam Style Questions
Question
If \(g(x)=x^{2}\sqrt{x+1}\), then \(g'(3)=\ ?\)
(A) \(\tfrac{3}{2}\)
(B) \(\tfrac{39}{4}\)
(C) \(\tfrac{57}{4}\)
(D) \(36\)
(B) \(\tfrac{39}{4}\)
(C) \(\tfrac{57}{4}\)
(D) \(36\)
▶️ Answer/Explanation
Correct answer: (C) \(\tfrac{57}{4}\)
Product rule and chain rule:
\(\displaystyle g'(x)=2x\sqrt{x+1}+x^{2}\cdot\frac{1}{2}(x+1)^{-1/2}\)
At \(x=3\): \(g'(3)=12+\tfrac{9}{4}=\tfrac{57}{4}\).
No-Calc Question
If \(f'(x)=x^{2}\) and \(g(x)=3e^{x}\) and \(h(x)=f(x)g(x)\), which of the following could be \(h'(x)\)?
(A) \(3x^{2}e^{x}\)
(B) \(x^{3}e^{x}+3x^{2}e^{x}\)
(C) \(x^{2}e^{x}+x^{2}e^{x}\)
(D) \(3x^{2}e^{x}+6xe^{x}\)
(B) \(x^{3}e^{x}+3x^{2}e^{x}\)
(C) \(x^{2}e^{x}+x^{2}e^{x}\)
(D) \(3x^{2}e^{x}+6xe^{x}\)
▶️ Answer/Explanation
Detailed solution
Product rule: \(h'(x)=f'(x)g(x)+f(x)g'(x)=x^{2}(3e^{x})+f(x)\cdot 3e^{x}\).
Since \(f'(x)=x^{2}\Rightarrow f(x)=\dfrac{x^{3}}{3}+C\), we get \(h'(x)=3x^{2}e^{x}+\big(\dfrac{x^{3}}{3}+C\big)3e^{x}=3x^{2}e^{x}+x^{3}e^{x}+3Ce^{x}\).
Dropping the unknown constant multiple still matches the form \(x^{3}e^{x}+3x^{2}e^{x}\).
✅ Answer: (B)