Math Core

Not every curve is the graph of a function $y = f(x)$ . A circle fails the vertical line test, and a curve like $x^3 + y^3 = 6xy$ can't be solved for $y$ in any reasonable way. Yet these curves still have tangent lines with slopes. Implicit differentiation finds those slopes without ever solving for $y$ .

Explicit versus implicit

The equation $y = x^2 - 3x$ defines $y$ explicitly: $y$ is alone on one side. The equation $x^2 + y^2 = 25$ defines $y$ implicitly: $x$ and $y$ are mixed together, and near most points the curve still behaves like the graph of some function of $x$ .

You could solve the circle for $y$ and get two functions, $y = \sqrt{25 - x^2}$ and $y = -\sqrt{25 - x^2}$ , then differentiate whichever half you need. Implicit differentiation is faster and works even when solving is impossible.

The key idea: $y$ is a function of $x$

Treat $y$ as an unknown function $y(x)$ . Then any expression involving $y$ is a composite function, and the chain rule applies. For example,

\frac{d}{dx}\big(y^2\big) = 2y \cdot \frac{dy}{dx}, \qquad \frac{d}{dx}\big(\sin y\big) = \cos y \cdot \frac{dy}{dx}, \qquad \frac{d}{dx}\big(e^y\big) = e^y \cdot \frac{dy}{dx}.

Compare with $\dfrac{d}{dx}(x^2) = 2x$ . The only difference is the factor $\dfrac{dy}{dx}$ , which is the inner derivative. Differentiating a $y$ term always produces a $\dfrac{dy}{dx}$ .

Implicit differentiation

Differentiate both sides of the equation with respect to $x$ . Every time you differentiate an expression containing $y$ , multiply by $\dfrac{dy}{dx}$ (chain rule). Use the product rule on terms like $xy$ .
Move every term containing $\dfrac{dy}{dx}$ to one side and everything else to the other.
Factor out $\dfrac{dy}{dx}$ and divide.

The result usually contains both $x$ and $y$ . To get a slope at a point, substitute both coordinates.

Worked example: Tangent line to a circle

Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$ , then find the tangent line at $(3, 4)$ .

Solution. Differentiate both sides:

2x + 2y\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{x}{y}.

At $(3, 4)$ the slope is $-\dfrac{3}{4}$ , so the tangent line is $y - 4 = -\dfrac{3}{4}(x - 3)$ , or $y = -\dfrac{3}{4}x + \dfrac{25}{4}$ .

This matches geometry: the radius to $(3, 4)$ has slope $\dfrac{4}{3}$ , and a tangent to a circle is perpendicular to the radius.

The circle x² + y² = 25 with its tangent line at (3, 4), slope −3/4.Open in grapher →

Worked example: A curve you can't solve for y

Find $\dfrac{dy}{dx}$ for $x^3 + y^3 = 6xy$ , and find the slope at $(3, 3)$ .

Solution. The right side is a product, so use the product rule there.

\begin{aligned} 3x^2 + 3y^2\frac{dy}{dx} &= 6y + 6x\frac{dy}{dx} \\ 3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} &= 6y - 3x^2 \\ \frac{dy}{dx}\big(3y^2 - 6x\big) &= 6y - 3x^2 \\ \frac{dy}{dx} &= \frac{2y - x^2}{y^2 - 2x}. \end{aligned}

First check that $(3, 3)$ is on the curve: $27 + 27 = 54 = 6 \cdot 3 \cdot 3$ . ✓ The slope there is $\dfrac{6 - 9}{9 - 6} = -1$ , so the tangent line is $y = -x + 6$ .

The curve x³ + y³ = 6xy and its tangent line y = −x + 6 at (3, 3).Open in grapher →

Worked example: Chain rule and product rule together

Find $\dfrac{dy}{dx}$ if $\sin(xy) = x$ .

Solution. The outer function is sine and the inside is the product $xy$ :

\cos(xy) \cdot \left(y + x\frac{dy}{dx}\right) = 1.

Distribute, then isolate $\dfrac{dy}{dx}$ :

y\cos(xy) + x\cos(xy)\frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1 - y\cos(xy)}{x\cos(xy)}.

Common mistake

The most common error is forgetting the $\dfrac{dy}{dx}$ when differentiating a $y$ term, for example writing $\dfrac{d}{dx}(y^3) = 3y^2$ . A close second is differentiating $xy$ as $1 \cdot \dfrac{dy}{dx}$ instead of using the product rule: $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$ .

Horizontal and vertical tangents

When $\dfrac{dy}{dx}$ is a fraction $\dfrac{N}{D}$ in $x$ and $y$ :

The tangent is horizontal where $N = 0$ and $D \ne 0$ .
The tangent is vertical where $D = 0$ and $N \ne 0$ .

In either case, you must also use the original equation to find the actual points on the curve.

Worked example: Vertical tangents on a tilted ellipse

For the curve $x^2 - xy + y^2 = 3$ , find every point where the tangent line is vertical.

Solution. Differentiate: $2x - \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$ , so

\frac{dy}{dx} = \frac{y - 2x}{2y - x}.

Vertical tangents need $2y - x = 0$ , so $x = 2y$ . Substitute into the curve: $4y^2 - 2y^2 + y^2 = 3y^2 = 3$ , giving $y = \pm 1$ . The points are $(2, 1)$ and $(-2, -1)$ . At each, the numerator is $\mp 3 \ne 0$ , so both tangents really are vertical.

The curve x² − xy + y² = 3 has vertical tangent lines at (2, 1) and (−2, −1).Open in grapher →

Tip

Before you substitute a point, check that it actually lies on the curve. AP questions sometimes include this as a first step, and it catches copying errors.

Practice

Practice 1

Find $\dfrac{dy}{dx}$ for the ellipse $x^2 + 4y^2 = 16$ . Give your answer in terms of $x$ and $y$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Find the slope of the ellipse $x^2 + 4y^2 = 16$ at the point $(2, \sqrt{3})$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find $\dfrac{dy}{dx}$ if $xy^2 + x^3 = 10$ . Give your answer in terms of $x$ and $y$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Find $\dfrac{dy}{dx}$ if $e^y = x^2 + y$ . Give your answer in terms of $x$ and $y$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

The point $(3, 1)$ lies on the curve $x^2y + y^3 = 10$ . Find an equation of the tangent line at that point.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

(AP-style) If $xy + y^2 = 6$ , what is the value of $\dfrac{dy}{dx}$ at the point $(1, 2)$ ?

Practice 7

(AP-style) At which points does the curve $x^2 - xy + y^2 = 3$ have a horizontal tangent line?

Explicit versus implicit

The key idea: yyy is a function of xxx

Horizontal and vertical tangents

Practice

The key idea: $y$ is a function of $x$