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Higher-order partial derivatives practice problems

Second partial derivatives, Clairaut's theorem, and whether a function is harmonic. 19 problems with worked solutions; in 19 of them every equation is proved by a computer algebra system.

14 worked problems

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

Problem 10.612

medium
For \( \displaystyle f(x, y) = \ln{\left(x^{2} + y^{2} \right)} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \ln{\left(x^{2} + y^{2} \right)} = \frac{- 2 x^{2} + 2 y^{2}}{\left(x^{2} + y^{2}\right)^{2}} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \ln{\left(x^{2} + y^{2} \right)} = \frac{2 x^{2} - 2 y^{2}}{\left(x^{2} + y^{2}\right)^{2}} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \ln{\left(x^{2} + y^{2} \right)} = - \frac{4 x y}{\left(x^{2} + y^{2}\right)^{2}} \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \ln{\left(x^{2} + y^{2} \right)} = - \frac{4 x y}{\left(x^{2} + y^{2}\right)^{2}} \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \ln{\left(x^{2} + y^{2} \right)} + \frac{\partial^{2}}{\partial y^{2}} \ln{\left(x^{2} + y^{2} \right)} = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = \frac{2 \left(- x^{2} + y^{2}\right)}{\left(x^{2} + y^{2}\right)^{2}},\ f_{yy} = \frac{2 \left(x^{2} - y^{2}\right)}{\left(x^{2} + y^{2}\right)^{2}},\ f_{xy} = - \frac{4 x y}{\left(x^{2} + y^{2}\right)^{2}};\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.613

medium
For \( \displaystyle f(x, y) = x e^{y} + y \sin{\left(x \right)} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x e^{y} + y \sin{\left(x \right)}\right) = - y \sin{\left(x \right)} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x e^{y} + y \sin{\left(x \right)}\right) = x e^{y} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x e^{y} + y \sin{\left(x \right)}\right) = e^{y} + \cos{\left(x \right)} \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x e^{y} + y \sin{\left(x \right)}\right) = e^{y} + \cos{\left(x \right)} \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x e^{y} + y \sin{\left(x \right)}\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x e^{y} + y \sin{\left(x \right)}\right) = x e^{y} - y \sin{\left(x \right)} \]
    f_xx + f_yy: not 0, so f is not harmonic.✓ Proved
Answer \( f_{xx} = - y \sin{\left(x \right)},\ f_{yy} = x e^{y},\ f_{xy} = e^{y} + \cos{\left(x \right)};\ \text{not harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.614

medium
For \( \displaystyle f(x, y) = e^{2 x} \sin{\left(2 y \right)} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} e^{2 x} \sin{\left(2 y \right)} = 4 e^{2 x} \sin{\left(2 y \right)} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} e^{2 x} \sin{\left(2 y \right)} = - 4 e^{2 x} \sin{\left(2 y \right)} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} e^{2 x} \sin{\left(2 y \right)} = 4 e^{2 x} \cos{\left(2 y \right)} \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} e^{2 x} \sin{\left(2 y \right)} = 4 e^{2 x} \cos{\left(2 y \right)} \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} e^{2 x} \sin{\left(2 y \right)} + \frac{\partial^{2}}{\partial y^{2}} e^{2 x} \sin{\left(2 y \right)} = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = 4 e^{2 x} \sin{\left(2 y \right)},\ f_{yy} = - 4 e^{2 x} \sin{\left(2 y \right)},\ f_{xy} = 4 e^{2 x} \cos{\left(2 y \right)};\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.616

medium
For \( \displaystyle f(x, y) = \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \sqrt{2} e^{- x} \cos{\left(y + \frac{\pi}{4} \right)} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \left(\sin{\left(y \right)} - \cos{\left(y \right)}\right) e^{- x} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \sqrt{2} e^{- x} \sin{\left(y + \frac{\pi}{4} \right)} \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \sqrt{2} e^{- x} \sin{\left(y + \frac{\pi}{4} \right)} \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} + \frac{\partial^{2}}{\partial y^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = \sqrt{2} e^{- x} \cos{\left(y + \frac{\pi}{4} \right)},\ f_{yy} = \left(\sin{\left(y \right)} - \cos{\left(y \right)}\right) e^{- x},\ f_{xy} = \sqrt{2} e^{- x} \sin{\left(y + \frac{\pi}{4} \right)};\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.617

medium
For \( \displaystyle f(x, y) = x^{3} - 3 x y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} - 3 x y^{2}\right) = 6 x \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} - 3 x y^{2}\right) = - 6 x \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{3} - 3 x y^{2}\right) = - 6 y \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{3} - 3 x y^{2}\right) = - 6 y \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} - 3 x y^{2}\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} - 3 x y^{2}\right) = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = 6 x,\ f_{yy} = - 6 x,\ f_{xy} = - 6 y;\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.618

medium
For \( \displaystyle f(x, y) = x^{2} - x y - y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} - x y - y^{2}\right) = 2 \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} - x y - y^{2}\right) = -2 \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{2} - x y - y^{2}\right) = -1 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{2} - x y - y^{2}\right) = -1 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} - x y - y^{2}\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} - x y - y^{2}\right) = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = 2,\ f_{yy} = -2,\ f_{xy} = -1;\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.619

medium
For \( \displaystyle f(x, y) = x^{2} + 2 x y - y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} + 2 x y - y^{2}\right) = 2 \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} + 2 x y - y^{2}\right) = -2 \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{2} + 2 x y - y^{2}\right) = 2 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{2} + 2 x y - y^{2}\right) = 2 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} + 2 x y - y^{2}\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} + 2 x y - y^{2}\right) = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = 2,\ f_{yy} = -2,\ f_{xy} = 2;\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.620

medium
For \( \displaystyle f(x, y) = x^{2} - 2 x y - y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} - 2 x y - y^{2}\right) = 2 \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} - 2 x y - y^{2}\right) = -2 \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{2} - 2 x y - y^{2}\right) = -2 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{2} - 2 x y - y^{2}\right) = -2 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} - 2 x y - y^{2}\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} - 2 x y - y^{2}\right) = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = 2,\ f_{yy} = -2,\ f_{xy} = -2;\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.621

medium
For \( \displaystyle f(x, y) = x^{3} y^{2} + x y \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} + x y\right) = 6 x y^{2} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} + x y\right) = 2 x^{3} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{3} y^{2} + x y\right) = 6 x^{2} y + 1 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{3} y^{2} + x y\right) = 6 x^{2} y + 1 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} + x y\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} + x y\right) = 2 x \left(x^{2} + 3 y^{2}\right) \]
    f_xx + f_yy: not 0, so f is not harmonic.✓ Proved
Answer \( f_{xx} = 6 x y^{2},\ f_{yy} = 2 x^{3},\ f_{xy} = 6 x^{2} y + 1;\ \text{not harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.622

medium
For \( \displaystyle f(x, y) = x^{3} y^{2} - 3 x y \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} - 3 x y\right) = 6 x y^{2} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} - 3 x y\right) = 2 x^{3} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{3} y^{2} - 3 x y\right) = 6 x^{2} y - 3 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{3} y^{2} - 3 x y\right) = 6 x^{2} y - 3 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} - 3 x y\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} - 3 x y\right) = 2 x \left(x^{2} + 3 y^{2}\right) \]
    f_xx + f_yy: not 0, so f is not harmonic.✓ Proved
Answer \( f_{xx} = 6 x y^{2},\ f_{yy} = 2 x^{3},\ f_{xy} = 6 x^{2} y - 3;\ \text{not harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.623

medium
For \( \displaystyle f(x, y) = x^{3} y^{2} - x y \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} - x y\right) = 6 x y^{2} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} - x y\right) = 2 x^{3} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{3} y^{2} - x y\right) = 6 x^{2} y - 1 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{3} y^{2} - x y\right) = 6 x^{2} y - 1 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} - x y\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} - x y\right) = 2 x \left(x^{2} + 3 y^{2}\right) \]
    f_xx + f_yy: not 0, so f is not harmonic.✓ Proved
Answer \( f_{xx} = 6 x y^{2},\ f_{yy} = 2 x^{3},\ f_{xy} = 6 x^{2} y - 1;\ \text{not harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.624

medium
For \( \displaystyle f(x, y) = e^{x} \sin{\left(y \right)} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} e^{x} \sin{\left(y \right)} = e^{x} \sin{\left(y \right)} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} e^{x} \sin{\left(y \right)} = - e^{x} \sin{\left(y \right)} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} e^{x} \sin{\left(y \right)} = e^{x} \cos{\left(y \right)} \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} e^{x} \sin{\left(y \right)} = e^{x} \cos{\left(y \right)} \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} e^{x} \sin{\left(y \right)} + \frac{\partial^{2}}{\partial y^{2}} e^{x} \sin{\left(y \right)} = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = e^{x} \sin{\left(y \right)},\ f_{yy} = - e^{x} \sin{\left(y \right)},\ f_{xy} = e^{x} \cos{\left(y \right)};\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.625

medium
For \( \displaystyle f(x, y) = x^{2} + x y - y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} + x y - y^{2}\right) = 2 \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} + x y - y^{2}\right) = -2 \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{2} + x y - y^{2}\right) = 1 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{2} + x y - y^{2}\right) = 1 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{2} + x y - y^{2}\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{2} + x y - y^{2}\right) = 0 \]
    f_xx + f_yy: 0, so f is harmonic.✓ Proved
Answer \( f_{xx} = 2,\ f_{yy} = -2,\ f_{xy} = 1;\ \text{harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 10.627

medium
For \( \displaystyle f(x, y) = x^{3} y^{2} + 2 x y \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
  1. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} + 2 x y\right) = 6 x y^{2} \]
    f_xx.✓ Proved
  2. \[ \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} + 2 x y\right) = 2 x^{3} \]
    f_yy.✓ Proved
  3. \[ \frac{\partial^{2}}{\partial y\partial x} \left(x^{3} y^{2} + 2 x y\right) = 6 x^{2} y + 2 \]
    f_xy.✓ Proved
  4. \[ \frac{\partial^{2}}{\partial x\partial y} \left(x^{3} y^{2} + 2 x y\right) = 6 x^{2} y + 2 \]
    f_yx, the same (Clairaut).✓ Proved
  5. \[ \frac{\partial^{2}}{\partial x^{2}} \left(x^{3} y^{2} + 2 x y\right) + \frac{\partial^{2}}{\partial y^{2}} \left(x^{3} y^{2} + 2 x y\right) = 2 x \left(x^{2} + 3 y^{2}\right) \]
    f_xx + f_yy: not 0, so f is not harmonic.✓ Proved
Answer \( f_{xx} = 6 x y^{2},\ f_{yy} = 2 x^{3},\ f_{xy} = 6 x^{2} y + 2;\ \text{not harmonic} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 higher-order partial derivatives practice

For \( \displaystyle f(x, y) = e^{3 x} \sin{\left(3 y \right)} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
Problem 10.615medium✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{2} - 3 x y - y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
Problem 10.626medium✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{3} y^{2} + 3 x y \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
Problem 10.628medium✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{3} y^{2} - 2 x y \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
Problem 10.704medium✓ Nihil obstat
For \( \displaystyle f(x, y) = x^{2} + 3 x y - y^{2} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).
Problem 10.705medium✓ Nihil obstat