# Partial Derivatives Practice Problems with Answers

> 12 partial derivative practice problems with answers: first and second partials, gradients, directional derivatives, tangent planes and critical points.

Source: https://practicecalculus.com/practice-problems/partial-derivatives
Updated: 2026-11-06

$$ f_x = \frac{\partial f}{\partial x}, \quad \nabla f = \langle f_x, f_y\rangle, \quad D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 3
- **You need:** All the single-variable derivative rules, especially the [chain rule](/rules/chain-rule)
- **Main idea:** Differentiate in one variable and treat every other variable as a constant
- **Practice with new numbers:** [Partial derivatives lesson](/learn/calculus-3/partial-derivatives/partial-derivatives)

## Key takeaways

- For $f_x$, treat $y$ as a number. For $f_y$, treat $x$ as a number. Then use the usual rules.
- Mixed partials $f_{xy}$ and $f_{yx}$ are equal for nice functions, by [Clairaut's theorem](/theorems/clairauts-theorem). Use that as a check.
- The directional derivative is the gradient dotted with a unit vector. Normalize the direction first.

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:

$$f_x = \frac{\partial f}{\partial x}: \text{ differentiate in } x, \text{ treat } y \text{ as a constant}$$

For the ideas with pictures, read the [partial derivatives guide](/blog/partial-derivatives) and the [gradient guide](/blog/gradient-and-directional-derivatives).

## 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

| Problem | Answer |
| --- | --- |
| 1 | $f_x = 3x^2y^2 + 4y$, $f_y = 2x^3y + 4x$ |
| 2 | $f_x = ye^{xy}$, $f_y = xe^{xy}$ |
| 3 | $f_x = 2xy\cos(x^2y)$, $f_y = x^2\cos(x^2y)$ |
| 4 | $f_x = \dfrac{2x}{x^2 + y^2}$ |
| 5 | $f_x = \dfrac{1}{y}$, $f_y = -\dfrac{x}{y^2}$ |
| 6 | $f_z = xy + 2z$ |
| 7 | $f_{xx} = 2y^3$, $f_{yy} = 6x^2y$, $f_{xy} = 6xy^2$ |
| 8 | $f_x(1, 2) = 8$, $f_y(1, 2) = -1$ |
| 9 | $\langle 4, 4\rangle$ |
| 10 | $\frac{28}{5}$ |
| 11 | $z = 5 + 2(x - 1) + 4(y - 2)$, or $z = 2x + 4y - 5$ |
| 12 | $(2, -3)$, a local minimum with value $-13$ |

## 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 $x$:

$$f_x = ye^{xy}, \qquad f_y = xe^{xy}$$

**3.** Chain rule with inside $x^2y$:

$$f_x = \cos(x^2y)\cdot 2xy, \qquad f_y = \cos(x^2y)\cdot x^2$$

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

$$f_x = \frac{2x}{x^2 + y^2}$$

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

$$f_x = \frac{1}{y}, \qquad f_y = -xy^{-2} = -\frac{x}{y^2}$$

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

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

$$f_{xx} = 2y^3, \qquad f_{yy} = 6x^2y, \qquad f_{xy} = \frac{\partial}{\partial y}(2xy^3) = 6xy^2$$

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 $\left\langle\frac{3}{5}, \frac{4}{5}\right\rangle$:

$$D_{\mathbf{u}}f = \langle 4, 4\rangle\cdot\left\langle\frac{3}{5}, \frac{4}{5}\right\rangle = \frac{12}{5} + \frac{16}{5} = \frac{28}{5}$$

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

$$z = 5 + 2(x - 1) + 4(y - 2) = 2x + 4y - 5$$

**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](/learn/calculus-3/partial-derivatives/partial-derivatives), the [second partials lesson](/learn/calculus-3/partial-derivatives/second-partial-derivatives), the [tangent planes lesson](/learn/calculus-3/partial-derivatives/tangent-planes) and the [gradient lesson](/learn/calculus-3/partial-derivatives/gradient-and-directional-derivatives). For optimization with a constraint, move on to the [Lagrange multiplier problems](/practice-problems/lagrange-multipliers).

## Frequently asked questions

### How do you take a partial derivative?

Pick the variable you are differentiating with respect to, and treat every other variable as a constant. For $f = x^3y^2$, $f_x = 3x^2y^2$, because $y^2$ is just a constant multiplier.

### What does the gradient tell you?

The gradient $\nabla f$ points in the direction of fastest increase, and its length is that fastest rate. It is also perpendicular to level curves.

### Why does the direction vector need to be a unit vector?

The directional derivative measures change per unit distance. A longer vector would scale the answer by its length. In problem 10, using $\langle 3, 4\rangle$ instead of $\langle\frac{3}{5}, \frac{4}{5}\rangle$ gives an answer five times too big.