AP Calculus BC 3.1 The Chain Rule - Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
Detailed solution
Let \(u=3+5x^2\).
\(\dfrac{d}{dx}\,u^{1/4}=\tfrac14 u^{-3/4}\cdot u’\) (chain rule).
\(u’=10x\).
Therefore \( \dfrac{d}{dx}(3+5x^2)^{1/4}=\tfrac14(3+5x^2)^{-3/4}(10x)\).
✅ Correct: D
Question
Compute \( \displaystyle \frac{d}{dx}\Big((x^{2}-11x+24)^{7}\Big) \).
(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
Detailed solution
Let \(g(x)=x^{2}-11x+24\). Then the expression is \(g(x)^{7}\). Chain rule: \[ \frac{d}{dx}\big(g(x)^{7}\big)=7\,g(x)^{6}\times g'(x). \] Compute \(g'(x)=2x-11\).
Therefore \[ \frac{d}{dx}\Big((x^{2}-11x+24)^{7}\Big) =7(x^{2}-11x+24)^{6}\times(2x-11). \] ✅ Answer: (D)