Partial Derivatives Practice Problems with Answers

These partial derivative practice problems start with first partials and move on to second partials, gradients, directional derivatives, tangent planes and critical points. Try each one first. The answers come first, then full solutions.

The one idea behind every problem:

For the ideas with pictures, read the partial derivatives guide and the gradient guide.

Problems

First partials

1. f(x, y) = x^3y^2 + 4xy. Find f_x and f_y.
2. f(x, y) = e^{xy}. Find f_x and f_y.
3. f(x, y) = \sin(x^2y). Find f_x and f_y.
4. f(x, y) = \ln(x^2 + y^2). Find f_x.
5. f(x, y) = \dfrac{x}{y}. Find f_x and f_y.
6. f(x, y, z) = xyz + z^2. Find f_z.

Second partials and values

7. f(x, y) = x^2y^3. Find f_{xx}, f_{yy} and f_{xy}.
8. f(x, y) = x^2 + 3xy - y^2. Find f_x(1, 2) and f_y(1, 2).

Gradients and planes

9. Find \nabla f(2, 1) for f(x, y) = x^2y.
10. Find the directional derivative of f(x, y) = x^2y at (2, 1) in the direction of \langle 3, 4\rangle.
11. Find the tangent plane to z = x^2 + y^2 at (1, 2, 5).
12. Find and classify the critical point of f(x, y) = x^2 + y^2 - 4x + 6y.

Answers

Step-by-step solutions

1. For f_x, y^2 and 4y are constants: f_x = 3x^2y^2 + 4y. For f_y, x^3 and 4x are constants: f_y = 2x^3y + 4x.

2. Chain rule with inside xy. Its x-derivative is y and its y-derivative is x:

3. Chain rule with inside x^2y:

4. Derivative of \ln u is \frac{u'}{u}, and the x-derivative of x^2 + y^2 is 2x:

5. In x, \frac{x}{y} is x times the constant \frac{1}{y}. In y, it is x\cdot y^{-1}:

6. Treat x and y as constants: f_z = xy + 2z.

7. First, f_x = 2xy^3 and f_y = 3x^2y^2. Then:

Check: f_{yx} = \frac{\partial}{\partial x}(3x^2y^2) = 6xy^2. They match.

8. f_x = 2x + 3y, so f_x(1, 2) = 2 + 6 = 8. f_y = 3x - 2y, so f_y(1, 2) = 3 - 4 = -1.

9. f_x = 2xy and f_y = x^2. At (2, 1): \nabla f = \langle 4, 4\rangle.

10. The length of \langle 3, 4\rangle is 5, so the unit vector is \left\langle\frac{3}{5}, \frac{4}{5}\right\rangle:

11. f_x = 2x = 2 and f_y = 2y = 4 at (1, 2). The tangent plane is z = f(a, b) + f_x(x - a) + f_y(y - b):

12. Set f_x = 2x - 4 = 0 and f_y = 2y + 6 = 0: (2, -3). The second derivative test gives D = f_{xx}f_{yy} - f_{xy}^2 = 2\cdot 2 - 0 = 4 > 0 with f_{xx} > 0, so it is a local minimum. The value is 4 + 9 - 8 - 18 = -13.

Common mistakes on these problems

• Differentiating the constant variable. In problem 1, f_x of 4xy is 4y, not 4y + 4x.
• Dropping the chain rule. In problem 3, the factor 2xy comes from the inside x^2y.
• Using a direction that is not a unit vector. In problem 10, normalize \langle 3, 4\rangle first.
• Mixing up f_{xy} order. f_{xy} means differentiate in x first, then y. For smooth functions both orders agree, which makes a good check.

Keep practicing

For unlimited problems with instant feedback, try the partial derivatives lesson, the second partials lesson, the tangent planes lesson and the gradient lesson. For optimization with a constraint, move on to the Lagrange multiplier problems.