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Derivatives

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''.

Derivatives Updated 4 min read

At a glance

\frac{d}{dx}\big[f(y)\big] = f'(y)\,\frac{dy}{dx}
Problems
12, with answers and full solutions
Level
Calculus 1 and AP Calculus AB, Unit 3
You need
The chain rule and the 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
On this page
  1. Problems
  2. Answers
  3. Step-by-step solutions
  4. Common mistakes on these problems
  5. Keep practicing

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 or the implicit differentiation guide.

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

  1. \tan y = x
  2. e^y = x^2 + y
  3. \cos(xy) = y

Tangent lines and second derivatives

  1. Find the tangent line to x^2 + y^2 = 25 at (3, -4).
  2. Find the tangent line to x^2 + xy + y^2 = 7 at (1, 2).
  3. Find the points where x^2 + xy + y^2 = 7 has a horizontal tangent line.
  4. 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, 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. Implicit differentiation with respect to time is the main tool in related rates practice problems, and it is how the inverse trig derivatives are proved.

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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.

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