Implicit Differentiation

Implicit differentiation finds dy/dx when y is not written by itself, as in x^2 + y^2 = 25. Differentiate both sides with respect to x, treating y as a function of x. Each time you differentiate a y term, multiply by dy/dx (the chain rule). Then solve for dy/dx.

Before this lesson: Chain Rule, Product Rule and Literal Equations

Explicit and implicit equations

An equation like y = x^3 - 4x is explicit: y is already alone, so you can differentiate the right side directly. An equation like x^2 + y^2 = 25 is implicit: x and y are mixed together, and solving for y would give two functions, y = \pm\sqrt{25 - x^2}.

Implicit differentiation skips that solving step. You keep the equation as it is and differentiate both sides. The key idea is that y stands for some function of x, so the derivative of a y expression needs the chain rule:

\frac{d}{dx}\left[y^2\right] = 2y \cdot \frac{dy}{dx}, \qquad \frac{d}{dx}\left[x^2\right] = 2x

The x term gets no extra factor, because the derivative of x with respect to x is 1. We will often write y′ for dy/dx.

How to do implicit differentiation

1. Differentiate both sides of the equation with respect to x.
2. Each time you differentiate a term with y, multiply by y′. Use the product rule for terms like xy.
3. Move every term with y′ to one side and every other term to the other side.
4. Factor out y′.
5. Divide to solve for y′. To get a slope at a point, substitute both x and y.

The answer usually contains both x and y. That is expected: on a circle, two points can share the same x but have different slopes, so the slope depends on y too.

Implicit differentiation examples

Example 1: The slope of a circle

Find dy/dx for x^2 + y^2 = 25, and find the tangent line at (3, 4).

2x + 2y\,y' = 0

2y\,y' = -2x

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

y' = -\frac{3}{4}

y - 4 = -\frac{3}{4}(x - 3)

y = -\frac{3}{4}x + \frac{25}{4}

Answer: \frac{dy}{dx} = -\frac{x}{y}, and the tangent line at (3, 4) is y = -\frac{3}{4}x + \frac{25}{4}.

Example 2: A product of x and y

Find dy/dx for x^2y + y^3 = 10, then find the slope at (1, 2).

2xy + x^2y' + 3y^2y' = 0

x^2y' + 3y^2y' = -2xy

y'(x^2 + 3y^2) = -2xy

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

y' = \frac{-2(1)(2)}{1 + 3(4)} = -\frac{4}{13}

Answer: \frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}, and the slope at (1, 2) is -\frac{4}{13}. Check that the point is on the curve first: 1 \cdot 2 + 8 = 10.

Example 3: The folium of Descartes

Find dy/dx for x^3 + y^3 = 6xy, and find the slope at (3, 3).

3x^2 + 3y^2y' = 6y + 6xy'

3y^2y' - 6xy' = 6y - 3x^2

y'(3y^2 - 6x) = 6y - 3x^2

y' = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}

y' = \frac{6 - 9}{9 - 6} = -1

Answer: \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}, and the slope at (3, 3) is −1. Solving this equation for y would be very hard, so implicit differentiation is the only practical method.

Example 4: Finding horizontal tangent lines

Find every point on x^2 - xy + y^2 = 12 where the tangent line is horizontal.

2x - (y + xy') + 2yy' = 0

-xy' + 2yy' = y - 2x

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

y - 2x = 0 \Rightarrow y = 2x

x^2 - x(2x) + (2x)^2 = 12

3x^2 = 12

x^2 = 4 \Rightarrow x = \pm 2

Answer: the horizontal tangents are at (2, 4) and (−2, −4), where y = 2x. At both points the denominator 2y − x is 6 or −6, not 0, so the slope really is 0.

Example 5: Word problem with price and demand

A shop finds that the price p (in dollars) and weekly sales q of a product are related by pq + 2q^2 = 3000. Right now it sells 30 per week. Find dp/dq and explain what it means.

30p + 2(900) = 3000

30p = 1200 \Rightarrow p = 40

p + q\frac{dp}{dq} + 4q = 0

\frac{dp}{dq} = -\frac{p + 4q}{q}

\frac{dp}{dq} = -\frac{40 + 120}{30} = -\frac{16}{3}

-\frac{16}{3} \approx -5.33

Answer: \frac{dp}{dq} \approx -5.33. To sell one more item per week, the shop must lower the price by about $5.33.

Common mistakes

Related methods

Solving for y first

For the circle, you could write the top half as y = \sqrt{25 - x^2} and use the chain rule: y' = \frac{-x}{\sqrt{25 - x^2}}. Since \sqrt{25 - x^2} = y on the top half, that is the same as -\frac{x}{y}. Implicit differentiation gets there faster and covers both halves at once.

Second derivatives by implicit differentiation

Differentiate y′ again and replace any y′ you get. For the circle, y' = -\frac{x}{y} and the quotient rule gives y'' = -\frac{y - xy'}{y^2}. Substituting y' = -\frac{x}{y} and x^2 + y^2 = 25 simplifies this to y'' = -\frac{25}{y^3}. At (3, 4), y'' = -\frac{25}{64}.

Coming next

Implicit differentiation is how you find the derivatives of inverse functions, which is the next lesson, and it is the main tool in related rates problems.

Practice problems

1. Find dy/dx for x^2 + y^2 = 100, then find the slope at (6, 8).
2. Find dy/dx for y^2 = 4x.
3. Find the slope of x^2 + 3y^2 = 13 at (1, 2).
4. Find dy/dx for xy = 12 and the slope at (3, 4).
5. Find dy/dx for x^2y = 8 and the slope at (2, 2).
6. Find the slope of y^3 + y = x^2 + 1 at (1, 1).
7. Find the tangent line to x^2 + xy + y^2 = 7 at (1, 2).
8. Find the slope of \sqrt{x} + \sqrt{y} = 5 at (4, 9).
9. Find the points on x^2 + y^2 - 6x = 0 where the tangent line is vertical.
10. A gas at constant temperature follows PV = 600, where P is pressure and V is volume. Find dV/dP when P = 20 and V = 30, and explain its meaning.

2x + 2yy' = 0

y' = -\frac{x}{y}

y' = -\frac{6}{8} = -\frac{3}{4}

2yy' = 4

y' = \frac{4}{2y} = \frac{2}{y}

2x + 6yy' = 0

y' = -\frac{2x}{6y} = -\frac{x}{3y}

y' = -\frac{1}{3(2)} = -\frac{1}{6}

y + xy' = 0

y' = -\frac{y}{x}

y' = -\frac{4}{3}

2xy + x^2y' = 0

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

y' = -\frac{2(2)}{2} = -2

3y^2y' + y' = 2x

y'(3y^2 + 1) = 2x

y' = \frac{2x}{3y^2 + 1}

y' = \frac{2}{4} = \frac{1}{2}

2x + y + xy' + 2yy' = 0

y'(x + 2y) = -(2x + y)

y' = -\frac{2x + y}{x + 2y} = -\frac{4}{5}

y - 2 = -\frac{4}{5}(x - 1)

y = -\frac{4}{5}x + \frac{14}{5}

\frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0

y' = -\frac{\sqrt{y}}{\sqrt{x}}

y' = -\frac{3}{2}

2x + 2yy' - 6 = 0

y' = \frac{3 - x}{y}

y = 0

x^2 - 6x = 0

x(x - 6) = 0 \Rightarrow x = 0 \text{ or } x = 6

V + P\frac{dV}{dP} = 0

\frac{dV}{dP} = -\frac{V}{P}

\frac{dV}{dP} = -\frac{30}{20} = -1.5

Implicit differentiation FAQ

When do you use implicit differentiation?

Use it when y is not alone and is hard or impossible to isolate, as in x^3 + y^3 = 6xy. It also works on equations you could solve, like circles.

Why do you multiply by dy/dx?

Because y is a function of x. By the chain rule, the derivative of an expression in y is its derivative with respect to y times dy/dx.

Why does the answer have both x and y?

An implicit curve can pass through several points with the same x. The slope can differ at each one, so you need y to tell which point you mean.

How do you find horizontal and vertical tangents?

A horizontal tangent is where the numerator of dy/dx is 0. A vertical tangent is where the denominator is 0. Check each point on the original curve.

Summary

• Differentiate both sides with respect to x, treating y as a function of x.
• Every y term gets a factor of y′. Use the product rule on terms like xy.
• Collect the y′ terms, factor out y′, and divide.
• Substitute both coordinates to get a slope at a point.
• Horizontal tangent: numerator 0. Vertical tangent: denominator 0.

Previous lesson: Chain Rule

Next lesson: Derivatives of Inverse Functions

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