AP Calculus AB 3.2 Implicit Differentiation - MCQs - Exam Style Questions
No-Calc Question
What is the slope of the line tangent to the graph of \(3x-y^{2}-6y=11\) at the point where \(y=4\)?
(A) \(-\tfrac{5}{6}\)
(B) \(-\tfrac{4}{7}\)
(C) \(\tfrac{3}{8}\)
(D) \(\tfrac{3}{14}\)
(B) \(-\tfrac{4}{7}\)
(C) \(\tfrac{3}{8}\)
(D) \(\tfrac{3}{14}\)
▶️ Answer/Explanation
Differentiate implicitly: \(3-2y\,y’-6y’=0\).
Factor \(y’\): \((2y+6)y’=3 \Rightarrow y’=\dfrac{3}{2y+6}\).
At \(y=4\): \(y’=\dfrac{3}{8+6}=\dfrac{3}{14}\).
✅ Answer: (D)
Factor \(y’\): \((2y+6)y’=3 \Rightarrow y’=\dfrac{3}{2y+6}\).
At \(y=4\): \(y’=\dfrac{3}{8+6}=\dfrac{3}{14}\).
✅ Answer: (D)
No-Calc Question
If \(x+3y^{1/3}=y\), what is \(\dfrac{dy}{dx}\) at the point \((2,8)\) ?
(A) \(\dfrac{1}{3}\)
(B) \(\dfrac{3}{4}\)
(C) \(\dfrac{5}{4}\)
(D) \(\dfrac{4}{3}\)
(B) \(\dfrac{3}{4}\)
(C) \(\dfrac{5}{4}\)
(D) \(\dfrac{4}{3}\)
▶️ Answer/Explanation
Differentiate implicitly: \(1+3\cdot \tfrac{1}{3}y^{-2/3}y’=y’\).
So \(1+y^{-2/3}y’=y’\Rightarrow 1=y'(1-y^{-2/3})\).
At \(y=8\): \(y^{-2/3}=1/4\).
Then \(1=y'(1-1/4)=y’\cdot \tfrac{3}{4}\Rightarrow y’=\tfrac{4}{3}\).
✅ Answer: (D)