AP Calculus AB 3.1 The Chain Rule - MCQs - Exam Style Questions
No-Calc Question
Find \[ \frac{d}{dx}\Big( x^{2}-11x+24 \Big)^{7}. \]
(A) \(7(2x-11)^{6}\)
(B) \(7(2x-11)^{6}(x^{2}-11x+24)\)
(C) \(7(x^{2}-11x+24)^{6}\)
(D) \(7(x^{2}-11x+24)^{6}(2x-11)\)
(B) \(7(2x-11)^{6}(x^{2}-11x+24)\)
(C) \(7(x^{2}-11x+24)^{6}\)
(D) \(7(x^{2}-11x+24)^{6}(2x-11)\)
▶️ Answer/Explanation
Use chain rule with \(u=x^{2}-11x+24\). Then \[ \frac{d}{dx}\big(u^{7}\big)=7u^{6}\cdot u’, \qquad u’=2x-11. \] Hence \[ \frac{d}{dx}\Big( x^{2}-11x+24 \Big)^{7} =7\big(x^{2}-11x+24\big)^{6}(2x-11). \] ✅ Answer: (D)
No-Calc Question
If \(y=\sin(x^{3})\), what is \(\dfrac{d^{2}y}{dx^{2}}\)?
(A) \(3x^{2}\cos(x^{3})\)
(B) \(9x^{4}\cos^{2}(x^{3})\)
(C) \(6x\cos(x^{3})-9x^{4}\sin(x^{3})\)
(D) \(-6x\cos(x^{3})+3x^{2}\sin(x^{3})\)
(B) \(9x^{4}\cos^{2}(x^{3})\)
(C) \(6x\cos(x^{3})-9x^{4}\sin(x^{3})\)
(D) \(-6x\cos(x^{3})+3x^{2}\sin(x^{3})\)
▶️ Answer/Explanation
First derivative (chain rule): \(y’=\cos(x^{3})\cdot 3x^{2}=3x^{2}\cos(x^{3})\).
Second derivative (product + chain rules):
\(\displaystyle y”=\frac{d}{dx}\big(3x^{2}\cos(x^{3})\big)=6x\cos(x^{3})+3x^{2}\big(-\sin(x^{3})\cdot 3x^{2}\big)\).
Simplify: \(\displaystyle y”=6x\cos(x^{3})-9x^{4}\sin(x^{3})\).
✅ Answer: (C)
Second derivative (product + chain rules):
\(\displaystyle y”=\frac{d}{dx}\big(3x^{2}\cos(x^{3})\big)=6x\cos(x^{3})+3x^{2}\big(-\sin(x^{3})\cdot 3x^{2}\big)\).
Simplify: \(\displaystyle y”=6x\cos(x^{3})-9x^{4}\sin(x^{3})\).
✅ Answer: (C)