AP Calculus BC 3.2 Implicit Differentiation - MCQs - Exam Style Questions
Question
If \(\,x^{2}-xy-3y^{2}=6\,\), then \(\displaystyle \frac{dy}{dx}=\)
A) \(\displaystyle \frac{2x}{1+6y}\)
B) \(\displaystyle \frac{2x-y}{x+6y}\)
C) \(\displaystyle \frac{2x+y}{x+6y}\)
D) \(\displaystyle \frac{y-2x}{x+6y}\)
B) \(\displaystyle \frac{2x-y}{x+6y}\)
C) \(\displaystyle \frac{2x+y}{x+6y}\)
D) \(\displaystyle \frac{y-2x}{x+6y}\)
▶️ Answer/Explanation
Detailed solution
Differentiate implicitly: \[ \frac{d}{dx}\!\big(x^{2}-xy-3y^{2}\big)=2x-\!\big(x\,\tfrac{dy}{dx}+y\big)-6y\,\tfrac{dy}{dx}=0. \]
Rearranging gives \[ (2x-y)-(x+6y)\tfrac{dy}{dx}=0 \quad\Rightarrow\quad \frac{dy}{dx}=\frac{2x-y}{x+6y}. \]
✅ Correct: B) \(\displaystyle \frac{2x-y}{x+6y}\)
No-Calc Question
What is the slope of the line tangent to the curve \(\sqrt{x}+\sqrt{y}=2\) at the point \(\left(\tfrac{9}{4}, \tfrac{1}{4}\right)\)?
(A) \(-3\)
(B) \(-\tfrac{1}{3}\)
(C) \(1\)
(D) \(\tfrac{4}{3}\)
(B) \(-\tfrac{1}{3}\)
(C) \(1\)
(D) \(\tfrac{4}{3}\)
▶️ Answer/Explanation
Detailed solution
Start with the equation: \(\sqrt{x}+\sqrt{y}=2\). Differentiate implicitly with respect to \(x\):
\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx}=0. \]
Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}. \]
Substitute the point \(\left(\tfrac{9}{4},\tfrac{1}{4}\right)\):
\[ \frac{dy}{dx}=-\frac{\sqrt{1/4}}{\sqrt{9/4}}=-\frac{1/2}{3/2}=-\tfrac{1}{3}. \]
✅ Answer: (B)