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Implicit differentiation practice problems

dy/dx when y is tangled up with x in an equation. 70 problems with worked solutions; in 70 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 2.758

hard
The curve \( \displaystyle - 2 x + y + \sin{\left(x y \right)} = 3 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 - \sin{\left(1 \right)} \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 2 x + Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 1\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 1} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{2 - \cos{\left(1 \right)}}{1 - \cos{\left(1 \right)}} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 1}, \quad \left.\frac{dy}{dx}\right|_{(-1,1)} = \frac{2 - \cos{\left(1 \right)}}{1 - \cos{\left(1 \right)}} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.765

hard
The curve \( \displaystyle - 2 x + 4 y + \sin{\left(x y \right)} = \sin{\left(1 \right)} + 2 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ \sin{\left(1 \right)} + 2 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 2 x + 4 Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 4\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 4} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{2 - \cos{\left(1 \right)}}{\cos{\left(1 \right)} + 4} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 4}, \quad \left.\frac{dy}{dx}\right|_{(1,1)} = \frac{2 - \cos{\left(1 \right)}}{\cos{\left(1 \right)} + 4} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.807

hard
The curve \( \displaystyle - 2 x + y + \sin{\left(x y \right)} = -3 + \sin{\left(2 \right)} \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ -3 + \sin{\left(2 \right)} \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 2 x + Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 1\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 1} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{2 - \cos{\left(2 \right)}}{2 \cos{\left(2 \right)} + 1} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 1}, \quad \left.\frac{dy}{dx}\right|_{(2,1)} = \frac{2 - \cos{\left(2 \right)}}{2 \cos{\left(2 \right)} + 1} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.932

hard
The curve \( \displaystyle - 3 x + 4 y + \sin{\left(x y \right)} = 7 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 7 - \sin{\left(1 \right)} \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 3 x + 4 Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 4\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 3 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 3}{x \cos{\left(x y \right)} + 4} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{3 - \cos{\left(1 \right)}}{4 - \cos{\left(1 \right)}} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 3}{x \cos{\left(x y \right)} + 4}, \quad \left.\frac{dy}{dx}\right|_{(-1,1)} = \frac{3 - \cos{\left(1 \right)}}{4 - \cos{\left(1 \right)}} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.933

hard
The curve \( \displaystyle - 2 x + 4 y + \sin{\left(x y \right)} = -6 - \sin{\left(1 \right)} \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ -6 - \sin{\left(1 \right)} \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 2 x + 4 Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 4\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 4} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{\cos{\left(1 \right)} + 2}{\cos{\left(1 \right)} + 4} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 4}, \quad \left.\frac{dy}{dx}\right|_{(1,-1)} = \frac{\cos{\left(1 \right)} + 2}{\cos{\left(1 \right)} + 4} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.1046

hard
The curve \( \displaystyle - x + 2 y + \sin{\left(x y \right)} = 3 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 - \sin{\left(1 \right)} \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- x + 2 Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 2\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 1 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 1}{x \cos{\left(x y \right)} + 2} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{1 - \cos{\left(1 \right)}}{2 - \cos{\left(1 \right)}} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 1}{x \cos{\left(x y \right)} + 2}, \quad \left.\frac{dy}{dx}\right|_{(-1,1)} = \frac{1 - \cos{\left(1 \right)}}{2 - \cos{\left(1 \right)}} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.755

medium
The curve \( \displaystyle x^{3} + 2 x y + y^{3} = -2 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ -2 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(x^{3} + 2 x Y{\left(x \right)} + Y^{3}{\left(x \right)}\right) = 3 x^{2} + \left(2 x + 3 Y^{2}{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} + 2 Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 3 x^{2} - 2 y}{2 x + 3 y^{2}} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ - \frac{1}{5} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- 3 x^{2} - 2 y}{2 x + 3 y^{2}}, \quad \left.\frac{dy}{dx}\right|_{(1,-1)} = - \frac{1}{5} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.756

medium
The curve \( \displaystyle 3 x^{2} + 3 x y + 3 y^{2} = 3 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(3 x^{2} + 3 x Y{\left(x \right)} + 3 Y^{2}{\left(x \right)}\right) = 6 x + \left(3 x + 6 Y{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} + 3 Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 6 x - 3 y}{3 x + 6 y} = \frac{- 2 x - y}{x + 2 y} \]
    Solve for dy/dx: minus F_x over F_y.✓ Proved
  5. \[ - \frac{1}{2} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- 2 x - y}{x + 2 y}, \quad \left.\frac{dy}{dx}\right|_{(0,1)} = - \frac{1}{2} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.757

medium
The curve \( \displaystyle x^{2} y - 3 x + y^{3} = 1 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 1 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(x^{2} Y{\left(x \right)} - 3 x + Y^{3}{\left(x \right)}\right) = 2 x Y{\left(x \right)} + \left(x^{2} + 3 Y^{2}{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} - 3 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 2 x y + 3}{x^{2} + 3 y^{2}} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ 1 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- 2 x y + 3}{x^{2} + 3 y^{2}}, \quad \left.\frac{dy}{dx}\right|_{(0,1)} = 1 \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.759

medium
The curve \( \displaystyle - 2 x + 3 y + \sin{\left(x y \right)} = 3 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 2 x + 3 Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 3\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 3} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 3}, \quad \left.\frac{dy}{dx}\right|_{(0,1)} = \frac{1}{3} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.760

medium
The curve \( \displaystyle 3 x^{2} - 2 x y + 2 y^{2} = 3 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(3 x^{2} - 2 x Y{\left(x \right)} + 2 Y^{2}{\left(x \right)}\right) = 6 x + \left(- 2 x + 4 Y{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} - 2 Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 6 x + 2 y}{- 2 x + 4 y} = \frac{3 x - y}{x - 2 y} \]
    Solve for dy/dx: minus F_x over F_y.✓ Proved
  5. \[ -2 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{3 x - y}{x - 2 y}, \quad \left.\frac{dy}{dx}\right|_{(1,1)} = -2 \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.761

medium
The curve \( \displaystyle - 2 x + 3 y + \sin{\left(x y \right)} = -2 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ -2 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 2 x + 3 Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 3\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 3} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ 1 \cdot \frac{1}{2} = \frac{1}{2} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 2}{x \cos{\left(x y \right)} + 3}, \quad \left.\frac{dy}{dx}\right|_{(1,0)} = \frac{1}{2} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.762

medium
The curve \( \displaystyle - x + y + \sin{\left(x y \right)} = 2 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 2 - \sin{\left(1 \right)} \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- x + Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 1\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 1 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 1}{x \cos{\left(x y \right)} + 1} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ 1 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 1}{x \cos{\left(x y \right)} + 1}, \quad \left.\frac{dy}{dx}\right|_{(-1,1)} = 1 \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.763

medium
The curve \( \displaystyle x^{3} - 2 x y + y^{3} = 5 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 5 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(x^{3} - 2 x Y{\left(x \right)} + Y^{3}{\left(x \right)}\right) = 3 x^{2} + \left(- 2 x + 3 Y^{2}{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} - 2 Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 3 x^{2} + 2 y}{- 2 x + 3 y^{2}} = \frac{3 x^{2} - 2 y}{2 x - 3 y^{2}} \]
    Solve for dy/dx: minus F_x over F_y.✓ Proved
  5. \[ 10 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{3 x^{2} - 2 y}{2 x - 3 y^{2}}, \quad \left.\frac{dy}{dx}\right|_{(2,1)} = 10 \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.764

medium
The curve \( \displaystyle x^{2} y - 2 x + 3 y^{3} = -2 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ -2 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(x^{2} Y{\left(x \right)} - 2 x + 3 Y^{3}{\left(x \right)}\right) = 2 x Y{\left(x \right)} + \left(x^{2} + 9 Y^{2}{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} - 2 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 2 x y + 2}{x^{2} + 9 y^{2}} \]
    Solve for dy/dx: minus F_x over F_y.✓ Proved
  5. \[ 2 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{2 \left(- x y + 1\right)}{x^{2} + 9 y^{2}}, \quad \left.\frac{dy}{dx}\right|_{(1,0)} = 2 \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.768

medium
The curve \( \displaystyle - 4 x + y + \sin{\left(x y \right)} = 1 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 1 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(- 4 x + Y{\left(x \right)} + \sin{\left(x Y{\left(x \right)} \right)}\right) = \left(x \cos{\left(x Y{\left(x \right)} \right)} + 1\right) \frac{d}{d x} Y{\left(x \right)} + Y{\left(x \right)} \cos{\left(x Y{\left(x \right)} \right)} - 4 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- y \cos{\left(x y \right)} + 4}{x \cos{\left(x y \right)} + 1} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ 3 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- y \cos{\left(x y \right)} + 4}{x \cos{\left(x y \right)} + 1}, \quad \left.\frac{dy}{dx}\right|_{(0,1)} = 3 \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.769

medium
The curve \( \displaystyle 3 x^{2} - x y + 3 y^{2} = 3 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(3 x^{2} - x Y{\left(x \right)} + 3 Y^{2}{\left(x \right)}\right) = 6 x + \left(- x + 6 Y{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} - Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 6 x + y}{- x + 6 y} = \frac{6 x - y}{x - 6 y} \]
    Solve for dy/dx: minus F_x over F_y.✓ Proved
  5. \[ 1 \cdot \frac{1}{6} = \frac{1}{6} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{6 x - y}{x - 6 y}, \quad \left.\frac{dy}{dx}\right|_{(0,1)} = \frac{1}{6} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.800

medium
The curve \( \displaystyle 3 x^{2} + 2 x y + y^{2} = 1 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 1 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(3 x^{2} + 2 x Y{\left(x \right)} + Y^{2}{\left(x \right)}\right) = 6 x + \left(2 x + 2 Y{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} + 2 Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 6 x - 2 y}{2 x + 2 y} = \frac{- 3 x - y}{x + y} \]
    Solve for dy/dx: minus F_x over F_y.✓ Proved
  5. \[ -1 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- 3 x - y}{x + y}, \quad \left.\frac{dy}{dx}\right|_{(0,1)} = -1 \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.805

medium
The curve \( \displaystyle 4 x^{2} + 3 x y + 2 y^{2} = 3 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ 3 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(4 x^{2} + 3 x Y{\left(x \right)} + 2 Y^{2}{\left(x \right)}\right) = 8 x + \left(3 x + 4 Y{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} + 3 Y{\left(x \right)} \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 8 x - 3 y}{3 x + 4 y} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ 5 \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- 8 x - 3 y}{3 x + 4 y}, \quad \left.\frac{dy}{dx}\right|_{(1,-1)} = 5 \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.808

medium
The curve \( \displaystyle x^{2} y - x + y^{3} = -3 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
  1. \[ -3 \]
    The point is on the curve.Reviewed
  2. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
    Reviewed
  3. \[ \frac{d}{d x} \left(x^{2} Y{\left(x \right)} - x + Y^{3}{\left(x \right)}\right) = 2 x Y{\left(x \right)} + \left(x^{2} + 3 Y^{2}{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} - 1 \]
    Every y term picks up a factor dy/dx.✓ Proved
  4. \[ \frac{- 2 x y + 1}{x^{2} + 3 y^{2}} \]
    Solve for dy/dx: minus F_x over F_y.Reviewed
  5. \[ \frac{3}{4} \]
    At the point.Reviewed
Answer \( \frac{dy}{dx} = \frac{- 2 x y + 1}{x^{2} + 3 y^{2}}, \quad \left.\frac{dy}{dx}\right|_{(1,-1)} = \frac{3}{4} \)

✓ Nihil obstat Lines: 1 proved, 4 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More implicit differentiation practice

The curve \( \displaystyle x^{3} - 2 x y + y^{3} = 0 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.766medium✓ Nihil obstat
The curve \( \displaystyle 2 x^{2} - 3 x y + 2 y^{2} = 4 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.767medium✓ Nihil obstat
The curve \( \displaystyle x^{3} - 2 x y + y^{3} = 2 \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.801medium✓ Every equation proved
The curve \( \displaystyle x^{3} - 2 x y + y^{3} = 2 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.802medium✓ Every equation proved
The curve \( \displaystyle 2 x^{2} + 3 x y + 2 y^{2} = 16 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.803medium✓ Nihil obstat
The curve \( \displaystyle - 2 x + 2 y + \sin{\left(x y \right)} = -4 - \sin{\left(1 \right)} \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.804medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 4 x + 2 y^{3} = -1 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.806medium✓ Nihil obstat
The curve \( \displaystyle 4 x^{2} - 2 x y + 3 y^{2} = 3 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.809medium✓ Every equation proved
The curve \( \displaystyle x^{3} + 2 x y + y^{3} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.810medium✓ Nihil obstat
The curve \( \displaystyle 2 x^{2} + 3 x y + 4 y^{2} = 2 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.811medium✓ Nihil obstat
The curve \( \displaystyle x^{2} - x y + 3 y^{2} = 5 \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.812medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - x + 3 y^{3} = 5 \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.813medium✓ Nihil obstat
The curve \( \displaystyle x^{2} - 2 x y + 3 y^{2} = 6 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.814medium✓ Nihil obstat
The curve \( \displaystyle - 3 x + 3 y + \sin{\left(x y \right)} = -3 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.927medium✓ Every equation proved
The curve \( \displaystyle 2 x^{2} - 3 x y + 2 y^{2} = 1 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.928medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 4 x + y^{3} = -6 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.929medium✓ Nihil obstat
The curve \( \displaystyle x^{2} - 2 x y + 3 y^{2} = 2 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.930medium✓ Every equation proved
The curve \( \displaystyle x^{2} y - 2 x + 2 y^{3} = 1 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.931medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 4 x + 4 y^{3} = -9 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.934medium✓ Nihil obstat
The curve \( \displaystyle - x + 2 y + \sin{\left(x y \right)} = 2 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1045medium✓ Nihil obstat
The curve \( \displaystyle x^{3} - 2 x y + y^{3} = 1 \) passes through \( \displaystyle (0, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1047medium✓ Every equation proved
The curve \( \displaystyle x^{3} + 3 x y + y^{3} = -3 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1048medium✓ Nihil obstat
The curve \( \displaystyle x^{3} - x y + y^{3} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1049medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 4 x + 4 y^{3} = 0 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1050medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 2 x + y^{3} = 1 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1051medium✓ Every equation proved
The curve \( \displaystyle x^{3} + 3 x y + y^{3} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1052medium✓ Nihil obstat
The curve \( \displaystyle x^{2} - 2 x y + y^{2} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1157medium✓ Every equation proved
The curve \( \displaystyle x^{3} + 2 x y + y^{3} = 4 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1158medium✓ Nihil obstat
The curve \( \displaystyle - 2 x + y + \sin{\left(x y \right)} = -2 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1159medium✓ Nihil obstat
The curve \( \displaystyle 4 x^{2} + 3 x y + 4 y^{2} = 26 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1160medium✓ Every equation proved
The curve \( \displaystyle x^{2} + 3 x y + 3 y^{2} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1161medium✓ Nihil obstat
The curve \( \displaystyle x^{3} + x y + y^{3} = -1 \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1162medium✓ Every equation proved
The curve \( \displaystyle x^{2} y - 2 x + y^{3} = 0 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1163medium✓ Nihil obstat
The curve \( \displaystyle x^{2} + 2 x y + 4 y^{2} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1164medium✓ Every equation proved
The curve \( \displaystyle x^{3} + 2 x y + y^{3} = 13 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1272medium✓ Nihil obstat
The curve \( \displaystyle x^{3} - x y + y^{3} = 1 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1275medium✓ Nihil obstat
The curve \( \displaystyle x^{3} + x y + y^{3} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1276medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 4 x + y^{3} = -3 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1277medium✓ Every equation proved
The curve \( \displaystyle 2 x^{2} + x y + y^{2} = 11 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1278medium✓ Every equation proved
The curve \( \displaystyle x^{2} y - 3 x + 3 y^{3} = 1 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1279medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 3 x + 4 y^{3} = 8 \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1468medium✓ Nihil obstat
The curve \( \displaystyle x^{2} y - 3 x + 2 y^{3} = 6 \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1469medium✓ Nihil obstat
The curve \( \displaystyle - 3 x + 3 y + \sin{\left(x y \right)} = 6 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1470medium✓ Every equation proved
The curve \( \displaystyle x^{2} y - x + 3 y^{3} = -1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1471medium✓ Nihil obstat
The curve \( \displaystyle 2 x^{2} + x y + 3 y^{2} = 13 \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1472medium✓ Every equation proved
The curve \( \displaystyle 4 x^{2} - 2 x y + y^{2} = 4 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1473medium✓ Every equation proved
The curve \( \displaystyle - 4 x + 4 y + \sin{\left(x y \right)} = -4 + \sin{\left(2 \right)} \) passes through \( \displaystyle (2, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1273hard✓ Every equation proved
The curve \( \displaystyle - x + 2 y + \sin{\left(x y \right)} = \sin{\left(1 \right)} + 1 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1274hard✓ Every equation proved
The curve \( \displaystyle - 2 x + 3 y + \sin{\left(x y \right)} = 5 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1474hard✓ Nihil obstat
The curve \( \displaystyle - 4 x + 2 y + \sin{\left(x y \right)} = 6 - \sin{\left(1 \right)} \) passes through \( \displaystyle (-1, 1) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
Problem 2.1475hard✓ Every equation proved