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AP Calculus BC : 3.2 Implicit Differentiation- Exam Style questions with Answer- MCQ

AP Calculus BC-Questions and Answers - All Topics
Question

Given that \( 3x – \tan y = 4 \), what is \( \frac{dy}{dx} \) in terms of \( y \)?

A) \( \frac{dy}{dx} = 3 \sin^2 y \)

B) \( \frac{dy}{dx} = 3 \cos^2 y \)

C) \( \frac{dy}{dx} = 3 \cos y \cot y \)

D) \( \frac{dy}{dx} = \frac{3}{1 + 9y^2} \)

▶️ Answer/Explanation
Solution
Given \( 3x – \tan y = 4 \), differentiate implicitly with respect to \( x \): \( \frac{d}{dx}(3x) – \frac{d}{dx}(\tan y) = \frac{d}{dx}(4) \).
This gives \( 3 – \sec^2 y \cdot \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \): \( \sec^2 y \cdot \frac{dy}{dx} = 3 \), so \( \frac{dy}{dx} = \frac{3}{\sec^2 y} \).
Since \( \sec^2 y = \frac{1}{\cos^2 y} \), then \( \frac{dy}{dx} = 3 \cos^2 y \).
✅ Answer: B)
Question

What is the slope of the line tangent to the curve \( 4y^2 + xy – 2x^2 = 3 \) at the point \( (-1, -1) \)?

A) \( -5 \)

B) \( -\frac{3}{7} \)

C) \( \frac{1}{4} \)

D) \( \frac{1}{3} \)

▶️ Answer/Explanation
Solution
Use implicit differentiation on \( 4y^2 + xy – 2x^2 = 3 \).
Differentiate: \( 8y \frac{dy}{dx} + (x \frac{dy}{dx} + y) – 4x = 0 \).
Simplify: \( (8y + x) \frac{dy}{dx} + y – 4x = 0 \), so \( \frac{dy}{dx} = \frac{4x – y}{8y + x} \).
Evaluate at \( (-1, -1) \): \( \frac{dy}{dx} = \frac{4(-1) – (-1)}{8(-1) + (-1)} = \frac{-3}{-9} = \frac{1}{3} \).
✅ Answer: D)
Question

If \( \cos(4x – y) = x + y \), then \( \frac{dy}{dx} = \)

A) \( -1 – \sin(4x – y) \)

B) \( \frac{2 + 4 \sin(4x – y)}{\sin(4x – y)} \)

C) \( \frac{-1}{1 + \sin(4x – y)} \)

D) \( \frac{1 + 4 \sin(4x – y)}{-1 + \sin(4x – y)} \)

▶️ Answer/Explanation
Solution
Differentiate \( \cos(4x – y) = x + y \) implicitly with respect to \( x \).
Left side: \( -\sin(4x – y) \cdot (4 – \frac{dy}{dx}) \).
Right side: \( 1 + \frac{dy}{dx} \).
Set equal: \( -\sin(4x – y) \cdot (4 – \frac{dy}{dx}) = 1 + \frac{dy}{dx} \).
Distribute: \( -4 \sin(4x – y) + \sin(4x – y) \cdot \frac{dy}{dx} = 1 + \frac{dy}{dx} \).
Solve: \( (\sin(4x – y) – 1) \frac{dy}{dx} = 1 + 4 \sin(4x – y) \).
Thus, \( \frac{dy}{dx} = \frac{1 + 4 \sin(4x – y)}{\sin(4x – y) – 1} = \frac{1 + 4 \sin(4x – y)}{-1 + \sin(4x – y)} \).
✅ Answer: D)
Question
\( f(1) = 4 \)\( f'(1) = -2 \)\( g(3) = 7 \)\( g'(3) = 1 \)

The point (1,3) lies on the curve in the xy-plane given by the equation \( f(x)g(y) = 24 + x + y \), where f is a differentiable function of x and g is a differentiable function of y. What is the value of \( \frac{dy}{dx} \) at the point (1,3)?

A) −11

B) 4

C) 5

D) 13
▶️Answer/Explanation
Markscheme

Answer: C

Solution:

1. Differentiate implicitly:

\[ \frac{d}{dx}[f(x)g(y)] = \frac{d}{dx}[24 + x + y] \]

\[ f'(x)g(y) + f(x)g'(y)\frac{dy}{dx} = 1 + \frac{dy}{dx} \]

2. Solve for dy/dx:

\[ f'(x)g(y) + f(x)g'(y)\frac{dy}{dx} – \frac{dy}{dx} = 1 \]

\[ \frac{dy}{dx}(f(x)g'(y) – 1) = 1 – f'(x)g(y) \]

\[ \frac{dy}{dx} = \frac{1 – f'(x)g(y)}{f(x)g'(y) – 1} \]

3. Substitute values at (1,3):

From the table:

  • \( f(1) = 4 \)
  • \( f'(1) = -2 \)
  • \( g(3) = 7 \)
  • \( g'(3) = 1 \)

4. Calculate:

\[ \frac{dy}{dx} = \frac{1 – (-2)(7)}{4(1) – 1} = \frac{1 + 14}{4 – 1} = \frac{15}{3} = 5 \]

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