# Implicit Differentiation Practice Problems with Answers

> 12 implicit differentiation practice problems with answers and full solutions: circles, products, trig and exponentials, tangent lines and y''.

Source: https://practicecalculus.com/practice-problems/implicit-differentiation
Updated: 2026-11-06

$$ \frac{d}{dx}\big[f(y)\big] = f'(y)\,\frac{dy}{dx} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 3
- **You need:** The [chain rule](/practice-problems/chain-rule) and the [product rule](/rules/product-rule)
- **Main idea:** Differentiate both sides with respect to x, attach dy/dx to every y term, then solve for dy/dx
- **Practice with new numbers:** [Implicit differentiation lesson](/learn/calculus-1/derivative-rules/implicit-differentiation)

## Key takeaways

- Every time you differentiate a term with $y$ in it, multiply by $\frac{dy}{dx}$. That is the chain rule at work.
- A term like $xy$ needs the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$.
- After differentiating, move every $\frac{dy}{dx}$ term to one side, factor it out and divide.

These **implicit differentiation practice problems** start with circles and simple products, then add trig and exponentials, and finish with tangent lines, horizontal tangents and a second derivative. Try each one first. The answers come right after the list, then the full solutions.

The key fact: since $y$ depends on $x$, the chain rule gives

$$\frac{d}{dx}\big[f(y)\big] = f'(y)\,\frac{dy}{dx}$$

So $\frac{d}{dx}y^2 = 2y\frac{dy}{dx}$ and $\frac{d}{dx}\sin y = \cos y\frac{dy}{dx}$. Below, $y'$ means $\frac{dy}{dx}$. For a refresher, see the [implicit differentiation rule](/rules/implicit-differentiation) or the [implicit differentiation guide](/blog/implicit-differentiation).

## Problems

**Find dy/dx**

1. $x^2 + y^2 = 25$
2. $xy = 8$
3. $\sqrt{x} + \sqrt{y} = 5$
4. $x^2 + xy + y^2 = 7$
5. $x^3 + y^3 = 6xy$

**Trig and exponentials**

6. $\tan y = x$
7. $e^y = x^2 + y$
8. $\cos(xy) = y$

**Tangent lines and second derivatives**

9. Find the tangent line to $x^2 + y^2 = 25$ at $(3, -4)$.
10. Find the tangent line to $x^2 + xy + y^2 = 7$ at $(1, 2)$.
11. Find the points where $x^2 + xy + y^2 = 7$ has a horizontal tangent line.
12. Find $y''$ for $x^2 + y^2 = 25$, in terms of $y$ only.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $-\dfrac{x}{y}$ |
| 2 | $-\dfrac{y}{x}$ |
| 3 | $-\dfrac{\sqrt{y}}{\sqrt{x}}$ |
| 4 | $-\dfrac{2x + y}{x + 2y}$ |
| 5 | $\dfrac{2y - x^2}{y^2 - 2x}$ |
| 6 | $\dfrac{1}{1 + x^2}$ |
| 7 | $\dfrac{2x}{e^y - 1}$ |
| 8 | $-\dfrac{y\sin(xy)}{1 + x\sin(xy)}$ |
| 9 | $y = \dfrac{3}{4}x - \dfrac{25}{4}$ |
| 10 | $y = -\dfrac{4}{5}x + \dfrac{14}{5}$ |
| 11 | $\left(\dfrac{\sqrt{21}}{3}, -\dfrac{2\sqrt{21}}{3}\right)$ and $\left(-\dfrac{\sqrt{21}}{3}, \dfrac{2\sqrt{21}}{3}\right)$ |
| 12 | $-\dfrac{25}{y^3}$ |

## Step-by-step solutions

**1.** Differentiate both sides. The right side is a constant:

$$2x + 2y\,y' = 0 \quad\Rightarrow\quad y' = -\frac{x}{y}$$

**2.** The left side is a product, so use the product rule:

$$1\cdot y + x\,y' = 0 \quad\Rightarrow\quad y' = -\frac{y}{x}$$

**3.** Write the roots as powers of $\frac{1}{2}$:

$$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\,y' = 0 \quad\Rightarrow\quad y' = -\frac{2\sqrt{y}}{2\sqrt{x}} = -\frac{\sqrt{y}}{\sqrt{x}}$$

**4.** The middle term needs the product rule:

$$2x + (y + x\,y') + 2y\,y' = 0$$

Collect the $y'$ terms: $y'(x + 2y) = -(2x + y)$, so

$$y' = -\frac{2x + y}{x + 2y}$$

**5.** The right side $6xy$ needs the product rule:

$$3x^2 + 3y^2\,y' = 6y + 6x\,y'$$

Move the $y'$ terms left and the rest right: $y'(3y^2 - 6x) = 6y - 3x^2$. Divide by $3$:

$$y' = \frac{2y - x^2}{y^2 - 2x}$$

**6.** Differentiate both sides: $\sec^2 y\cdot y' = 1$, so $y' = \frac{1}{\sec^2 y}$. Use $\sec^2 y = 1 + \tan^2 y = 1 + x^2$:

$$y' = \frac{1}{1 + x^2}$$

This is exactly how you prove the [derivative of arctan x](/derivatives/arctan-x), since $\tan y = x$ means $y = \arctan x$.

**7.** Differentiate: $e^y\,y' = 2x + y'$. Collect $y'$:

$$y'(e^y - 1) = 2x \quad\Rightarrow\quad y' = \frac{2x}{e^y - 1}$$

**8.** The left side is a chain rule with a product inside:

$$-\sin(xy)\cdot(y + x\,y') = y'$$

Expand: $-y\sin(xy) - x\sin(xy)\,y' = y'$. Collect $y'$ on the right:

$$-y\sin(xy) = y'\big(1 + x\sin(xy)\big) \quad\Rightarrow\quad y' = -\frac{y\sin(xy)}{1 + x\sin(xy)}$$

**9.** From problem 1, $y' = -\frac{x}{y}$. At $(3, -4)$ the slope is $-\frac{3}{-4} = \frac{3}{4}$:

$$y + 4 = \frac{3}{4}(x - 3) \quad\Rightarrow\quad y = \frac{3}{4}x - \frac{25}{4}$$

**10.** First check the point: $1 + 2 + 4 = 7$. From problem 4, the slope at $(1, 2)$ is

$$y' = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5}$$

$$y - 2 = -\frac{4}{5}(x - 1) \quad\Rightarrow\quad y = -\frac{4}{5}x + \frac{14}{5}$$

**11.** A horizontal tangent needs $y' = 0$, so the top of the answer to problem 4 must be zero: $2x + y = 0$, which means $y = -2x$. Substitute into the curve:

$$x^2 + x(-2x) + (-2x)^2 = 3x^2 = 7 \quad\Rightarrow\quad x = \pm\sqrt{\tfrac{7}{3}} = \pm\frac{\sqrt{21}}{3}$$

Then $y = -2x$ gives the two points in the answer table. At both points the bottom $x + 2y = -3x$ is not zero, so the slope really is $0$.

**12.** Start from $y' = -\frac{x}{y}$ and use the quotient rule:

$$y'' = -\frac{1\cdot y - x\,y'}{y^2}$$

Substitute $y' = -\frac{x}{y}$:

$$y'' = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}$$

On the curve $x^2 + y^2 = 25$, so $y'' = -\frac{25}{y^3}$.

## Common mistakes on these problems

- **Forgetting $y'$ on $y$ terms.** $\frac{d}{dx}y^2$ is $2y\,y'$, not $2y$. Every $y$ term must carry the factor.
- **Treating $xy$ as one variable.** In problem 2, $\frac{d}{dx}(xy)$ needs the product rule, giving two terms.
- **Differentiating the constant as something.** In problem 1, the derivative of $25$ is $0$.
- **Not substituting $y'$ in a second derivative.** In problem 12, replace $y'$ with $-\frac{x}{y}$ before you simplify.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new curves and instant feedback, try the [implicit differentiation lesson](/learn/calculus-1/derivative-rules/implicit-differentiation). Implicit differentiation with respect to time is the main tool in [related rates practice problems](/practice-problems/related-rates), and it is how the [inverse trig derivatives](/practice-problems/inverse-trig-derivatives) are proved.

## Frequently asked questions

### When do you use implicit differentiation?

Use it when $y$ is mixed in with $x$ and you cannot easily solve for $y$, as in $x^3 + y^3 = 6xy$. It also works when you could solve for $y$, like on a circle, and is often faster.

### Why do you multiply by dy/dx in implicit differentiation?

Because $y$ is a function of $x$. Differentiating $y^2$ with respect to $x$ is a chain rule problem: the outside is a square and the inside is $y$, so the result is $2y\cdot\frac{dy}{dx}$.

### How do you find a tangent line with implicit differentiation?

Find $\frac{dy}{dx}$ implicitly, plug in both coordinates of the point to get the slope, then use point-slope form. Problems 9 and 10 show this.