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Partial derivatives practice problems

Differentiate in one variable while holding the others fixed. 63 problems with worked solutions; in 63 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 10.1

hard
For \( \displaystyle f(x, y) = \ln{\left(x^{2} + y^{2} + 1 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \ln{\left(x^{2} + y^{2} + 1 \right)} = \frac{2 x}{x^{2} + y^{2} + 1} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \ln{\left(x^{2} + y^{2} + 1 \right)} = \frac{2 y}{x^{2} + y^{2} + 1} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \frac{2 x}{x^{2} + y^{2} + 1} = - \frac{4 x y}{\left(x^{2} + y^{2} + 1\right)^{2}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{2 y}{x^{2} + y^{2} + 1} = - \frac{4 x y}{\left(x^{2} + y^{2} + 1\right)^{2}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{2 x}{x^{2} + y^{2} + 1},\quad f_y = \frac{2 y}{x^{2} + y^{2} + 1},\quad f_{xy} = - \frac{4 x y}{\left(x^{2} + y^{2} + 1\right)^{2}} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.2

hard
For \( \displaystyle f(x, y) = x^{3} - 2 x y + y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x^{3} - 2 x y + y^{2}\right) = 3 x^{2} - 2 y \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x^{3} - 2 x y + y^{2}\right) = - 2 x + 2 y \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(3 x^{2} - 2 y\right) = -2 \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(- 2 x + 2 y\right) = -2 \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 3 x^{2} - 2 y,\quad f_y = - 2 x + 2 y,\quad f_{xy} = -2 \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.3

hard
For \( \displaystyle f(x, y) = \frac{x + y}{x - y + 3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \frac{x + y}{x - y + 3} = \frac{3 - 2 y}{\left(x - y + 3\right)^{2}} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \frac{x + y}{x - y + 3} = \frac{2 x + 3}{\left(x - y + 3\right)^{2}} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \frac{3 - 2 y}{\left(x - y + 3\right)^{2}} = \frac{- 2 x - 2 y}{\left(x - y + 3\right)^{3}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{2 x + 3}{\left(x - y + 3\right)^{2}} = \frac{- 2 x - 2 y}{\left(x - y + 3\right)^{3}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{3 - 2 y}{\left(x - y + 3\right)^{2}},\quad f_y = \frac{2 x + 3}{\left(x - y + 3\right)^{2}},\quad f_{xy} = \frac{2 \left(- x - y\right)}{\left(x - y + 3\right)^{3}} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.4

hard
For \( \displaystyle f(x, y) = \sqrt{x^{2} + y^{2} + 4} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \sqrt{x^{2} + y^{2} + 4} = \frac{x}{\sqrt{x^{2} + y^{2} + 4}} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \sqrt{x^{2} + y^{2} + 4} = \frac{y}{\sqrt{x^{2} + y^{2} + 4}} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \frac{x}{\sqrt{x^{2} + y^{2} + 4}} = - \frac{x y}{\left(x^{2} + y^{2} + 4\right)^{\frac{3}{2}}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{y}{\sqrt{x^{2} + y^{2} + 4}} = - \frac{x y}{\left(x^{2} + y^{2} + 4\right)^{\frac{3}{2}}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{x}{\sqrt{x^{2} + y^{2} + 4}},\quad f_y = \frac{y}{\sqrt{x^{2} + y^{2} + 4}},\quad f_{xy} = - \frac{x y}{\left(x^{2} + y^{2} + 4\right)^{\frac{3}{2}}} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.5

hard
For \( \displaystyle f(x, y) = e^{x y} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} e^{x y} = y e^{x y} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} e^{x y} = x e^{x y} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} y e^{x y} = \left(x y + 1\right) e^{x y} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} x e^{x y} = \left(x y + 1\right) e^{x y} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = y e^{x y},\quad f_y = x e^{x y},\quad f_{xy} = \left(x y + 1\right) e^{x y} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.6

hard
For \( \displaystyle f(x, y) = x \sin{\left(y \right)} + y \cos{\left(x \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x \sin{\left(y \right)} + y \cos{\left(x \right)}\right) = - y \sin{\left(x \right)} + \sin{\left(y \right)} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x \sin{\left(y \right)} + y \cos{\left(x \right)}\right) = x \cos{\left(y \right)} + \cos{\left(x \right)} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(- y \sin{\left(x \right)} + \sin{\left(y \right)}\right) = - \sin{\left(x \right)} + \cos{\left(y \right)} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(x \cos{\left(y \right)} + \cos{\left(x \right)}\right) = - \sin{\left(x \right)} + \cos{\left(y \right)} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = - y \sin{\left(x \right)} + \sin{\left(y \right)},\quad f_y = x \cos{\left(y \right)} + \cos{\left(x \right)},\quad f_{xy} = - \sin{\left(x \right)} + \cos{\left(y \right)} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.7

hard
For \( \displaystyle f(x, y) = x^{2} y + 3 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x^{2} y + 3 x y^{3}\right) = y \left(2 x + 3 y^{2}\right) \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x^{2} y + 3 x y^{3}\right) = x \left(x + 9 y^{2}\right) \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} y \left(2 x + 3 y^{2}\right) = 2 x + 9 y^{2} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} x \left(x + 9 y^{2}\right) = 2 x + 9 y^{2} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = y \left(2 x + 3 y^{2}\right),\quad f_y = x \left(x + 9 y^{2}\right),\quad f_{xy} = 2 x + 9 y^{2} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.8

hard
For \( \displaystyle f(x, y) = x e^{2 y} + y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x e^{2 y} + y^{2}\right) = e^{2 y} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x e^{2 y} + y^{2}\right) = 2 x e^{2 y} + 2 y \]
    f_y.✓ Proved
  4. \[ \frac{d}{d y} e^{2 y} = 2 e^{2 y} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(2 x e^{2 y} + 2 y\right) = 2 e^{2 y} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = e^{2 y},\quad f_y = 2 x e^{2 y} + 2 y,\quad f_{xy} = 2 e^{2 y} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.123

hard
For \( \displaystyle f(x, y) = 5 x^{2} y - 4 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(5 x^{2} y - 4 x y^{3}\right) = 10 x y - 4 y^{3} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(5 x^{2} y - 4 x y^{3}\right) = x \left(5 x - 12 y^{2}\right) \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(10 x y - 4 y^{3}\right) = 10 x - 12 y^{2} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} x \left(5 x - 12 y^{2}\right) = 10 x - 12 y^{2} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 10 x y - 4 y^{3},\quad f_y = x \left(5 x - 12 y^{2}\right),\quad f_{xy} = 10 x - 12 y^{2} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.125

hard
For \( \displaystyle f(x, y) = \frac{x + 3 y}{x - y + 5} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \frac{x + 3 y}{x - y + 5} = \frac{5 - 4 y}{\left(x - y + 5\right)^{2}} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \frac{x + 3 y}{x - y + 5} = \frac{4 x + 15}{\left(x - y + 5\right)^{2}} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \frac{5 - 4 y}{\left(x - y + 5\right)^{2}} = \frac{- 4 x - 4 y - 10}{\left(x - y + 5\right)^{3}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{4 x + 15}{\left(x - y + 5\right)^{2}} = \frac{- 4 x - 4 y - 10}{\left(x - y + 5\right)^{3}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{5 - 4 y}{\left(x - y + 5\right)^{2}},\quad f_y = \frac{4 x + 15}{\left(x - y + 5\right)^{2}},\quad f_{xy} = \frac{2 \left(- 2 x - 2 y - 5\right)}{\left(x - y + 5\right)^{3}} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.126

hard
For \( \displaystyle f(x, y) = \ln{\left(x^{2} + 3 y^{2} + 1 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \ln{\left(x^{2} + 3 y^{2} + 1 \right)} = \frac{2 x}{x^{2} + 3 y^{2} + 1} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \ln{\left(x^{2} + 3 y^{2} + 1 \right)} = \frac{6 y}{x^{2} + 3 y^{2} + 1} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \frac{2 x}{x^{2} + 3 y^{2} + 1} = - \frac{12 x y}{\left(x^{2} + 3 y^{2} + 1\right)^{2}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{6 y}{x^{2} + 3 y^{2} + 1} = - \frac{12 x y}{\left(x^{2} + 3 y^{2} + 1\right)^{2}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{2 x}{x^{2} + 3 y^{2} + 1},\quad f_y = \frac{6 y}{x^{2} + 3 y^{2} + 1},\quad f_{xy} = - \frac{12 x y}{\left(x^{2} + 3 y^{2} + 1\right)^{2}} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.127

hard
For \( \displaystyle f(x, y) = 2 \operatorname{atan}{\left(\frac{y}{x} \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} 2 \operatorname{atan}{\left(\frac{y}{x} \right)} = - \frac{2 y}{x^{2} + y^{2}} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} 2 \operatorname{atan}{\left(\frac{y}{x} \right)} = \frac{2 x}{x^{2} + y^{2}} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(- \frac{2 y}{x^{2} + y^{2}}\right) = \frac{- 2 x^{2} + 2 y^{2}}{\left(x^{2} + y^{2}\right)^{2}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{2 x}{x^{2} + y^{2}} = \frac{- 2 x^{2} + 2 y^{2}}{\left(x^{2} + y^{2}\right)^{2}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = - \frac{2 y}{x^{2} + y^{2}},\quad f_y = \frac{2 x}{x^{2} + y^{2}},\quad f_{xy} = \frac{2 \left(- x^{2} + y^{2}\right)}{\left(x^{2} + y^{2}\right)^{2}} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.130

hard
For \( \displaystyle f(x, y) = 2 x^{2} y + 4 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(2 x^{2} y + 4 x y^{3}\right) = 4 y \left(x + y^{2}\right) \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(2 x^{2} y + 4 x y^{3}\right) = 2 x \left(x + 6 y^{2}\right) \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} 4 y \left(x + y^{2}\right) = 4 x + 12 y^{2} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} 2 x \left(x + 6 y^{2}\right) = 4 x + 12 y^{2} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 4 y \left(x + y^{2}\right),\quad f_y = 2 x \left(x + 6 y^{2}\right),\quad f_{xy} = 4 x + 12 y^{2} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.131

hard
For \( \displaystyle f(x, y) = x^{3} - 3 x y + 3 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x^{3} - 3 x y + 3 y^{2}\right) = 3 x^{2} - 3 y \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x^{3} - 3 x y + 3 y^{2}\right) = - 3 x + 6 y \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(3 x^{2} - 3 y\right) = -3 \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(- 3 x + 6 y\right) = -3 \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 3 x^{2} - 3 y,\quad f_y = - 3 x + 6 y,\quad f_{xy} = -3 \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.133

hard
For \( \displaystyle f(x, y) = e^{2 x y} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} e^{2 x y} = 2 y e^{2 x y} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} e^{2 x y} = 2 x e^{2 x y} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} 2 y e^{2 x y} = \left(4 x y + 2\right) e^{2 x y} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} 2 x e^{2 x y} = \left(4 x y + 2\right) e^{2 x y} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 2 y e^{2 x y},\quad f_y = 2 x e^{2 x y},\quad f_{xy} = \left(4 x y + 2\right) e^{2 x y} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.136

hard
For \( \displaystyle f(x, y) = x^{3} + x y + 3 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x^{3} + x y + 3 y^{2}\right) = 3 x^{2} + y \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x^{3} + x y + 3 y^{2}\right) = x + 6 y \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(3 x^{2} + y\right) = 1 \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(x + 6 y\right) = 1 \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 3 x^{2} + y,\quad f_y = x + 6 y,\quad f_{xy} = 1 \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.137

hard
For \( \displaystyle f(x, y) = x^{2} \ln{\left(y + 1 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} x^{2} \ln{\left(y + 1 \right)} = 2 x \ln{\left(y + 1 \right)} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} x^{2} \ln{\left(y + 1 \right)} = \frac{x^{2}}{y + 1} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} 2 x \ln{\left(y + 1 \right)} = \frac{2 x}{y + 1} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{x^{2}}{y + 1} = \frac{2 x}{y + 1} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = 2 x \ln{\left(y + 1 \right)},\quad f_y = \frac{x^{2}}{y + 1},\quad f_{xy} = \frac{2 x}{y + 1} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.139

hard
For \( \displaystyle f(x, y) = x e^{y} - 2 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x e^{y} - 2 y^{2}\right) = e^{y} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x e^{y} - 2 y^{2}\right) = x e^{y} - 4 y \]
    f_y.✓ Proved
  4. \[ \frac{d}{d y} e^{y} = e^{y} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(x e^{y} - 4 y\right) = e^{y} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = e^{y},\quad f_y = x e^{y} - 4 y,\quad f_{xy} = e^{y} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.144

hard
For \( \displaystyle f(x, y) = x e^{3 y} - 2 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(x e^{3 y} - 2 y^{2}\right) = e^{3 y} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(x e^{3 y} - 2 y^{2}\right) = 3 x e^{3 y} - 4 y \]
    f_y.✓ Proved
  4. \[ \frac{d}{d y} e^{3 y} = 3 e^{3 y} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(3 x e^{3 y} - 4 y\right) = 3 e^{3 y} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = e^{3 y},\quad f_y = 3 x e^{3 y} - 4 y,\quad f_{xy} = 3 e^{3 y} \)

✓ Nihil obstat Lines: 4 proved, 1 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 10.170

hard
For \( \displaystyle f(x, y) = 3 x \sin{\left(y \right)} + y \cos{\left(x \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
    Reviewed
  2. \[ \frac{\partial}{\partial x} \left(3 x \sin{\left(y \right)} + y \cos{\left(x \right)}\right) = - y \sin{\left(x \right)} + 3 \sin{\left(y \right)} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} \left(3 x \sin{\left(y \right)} + y \cos{\left(x \right)}\right) = 3 x \cos{\left(y \right)} + \cos{\left(x \right)} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(- y \sin{\left(x \right)} + 3 \sin{\left(y \right)}\right) = - \sin{\left(x \right)} + 3 \cos{\left(y \right)} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \left(3 x \cos{\left(y \right)} + \cos{\left(x \right)}\right) = - \sin{\left(x \right)} + 3 \cos{\left(y \right)} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = - y \sin{\left(x \right)} + 3 \sin{\left(y \right)},\quad f_y = 3 x \cos{\left(y \right)} + \cos{\left(x \right)},\quad f_{xy} = - \sin{\left(x \right)} + 3 \cos{\left(y \right)} \)

✓ Nihil obstat Lines: 4 proved, 1 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 partial derivatives practice

For \( \displaystyle f(x, y) = 4 x^{2} y - 3 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.124hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x^{2} y - 4 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.128hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \frac{x + 3 y}{x - y + 2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.129hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \ln{\left(x^{2} + 2 y^{2} + 1 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.132hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{3} - 3 x y + 2 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.134hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 4 x^{2} y - 4 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.135hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \sqrt{x^{2} + y^{2} + 1} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.138hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 5 x^{2} y + 3 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.140hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \frac{x + 2 y}{x - y + 1} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.141hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \frac{x + 3 y}{x - y + 3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.142hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x^{2} y - 2 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.143hard✓ Nihil obstat
For \( \displaystyle f(x, y) = e^{3 x y} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.145hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{2} y^{2} - 3 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.171hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x^{2} y + 2 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.172hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x^{3} y^{2} - y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.173hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x y^{2} - y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.174hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 2 x \sin{\left(3 y \right)} + y \cos{\left(x \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.175hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 5 x y^{2} - 2 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.176hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \operatorname{atan}{\left(\frac{y}{x} \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.177hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{3 y} + 2 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.203hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \frac{x + y}{x - y + 1} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.204hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{2} y + x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.205hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x^{2} y + 3 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.206hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{2 y} - 3 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.207hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x^{2} y - x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.208hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{y} + 2 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.209hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{y} + y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.210hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{3} + x y + 5 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.235hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \frac{x + y}{x - y + 5} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.236hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 4 x^{2} y + 2 x y^{3} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.237hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{2 y} + 4 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.238hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 2 x^{2} y^{2} - 3 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.239hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{2 y} - 4 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.240hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{2} \ln{\left(y + 3 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.241hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{2} y^{2} + 4 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.242hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \sqrt{x^{2} + y^{2} + 5} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.267hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 2 x y^{2} + 3 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.268hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 4 x^{3} y^{2} + 4 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.269hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{3 y} + 4 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.270hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{3 y} + 3 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.271hard✓ Nihil obstat
For \( \displaystyle f(x, y) = x e^{y} + 4 y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.272hard✓ Nihil obstat
For \( \displaystyle f(x, y) = 3 x y^{2} - 2 y \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.273hard✓ Nihil obstat
For \( \displaystyle f(x, y) = \frac{x + y}{x - y + 4} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
Problem 10.274hard✓ Every equation proved