Implicit Differentiation

Some curves, like the circle $x^2 + y^2 = 25$, aren’t written as ”$y =$ something”. Solving for $y$ can be messy or even impossible. Implicit differentiation lets you find $\dfrac{dy}{dx}$ anyway, by differentiating both sides of the equation as it stands. It’s really just the chain rule used in a clever way.

Key ideas

Explicit and implicit

An equation like $y = x^3 - 2x$ defines $y$ explicitly: $y$ is alone on one side. An equation like $x^2 + xy + y^2 = 7$ defines $y$ implicitly: $x$ and $y$ are mixed together. We still think of $y$ as a function of $x$ (at least near any given point), even though we don’t have a formula for it.

Differentiating terms with y

Since $y$ is a function of $x$, any expression in $y$ is a composite function, with $y$ as the inside. So the chain rule adds a factor of $\dfrac{dy}{dx}$:

Term	Derivative with respect to $x$
$x^3$	$3x^2$
$y^3$	$3y^2 \dfrac{dy}{dx}$
$\sin y$	$\cos y \dfrac{dy}{dx}$
$e^y$	$e^y \dfrac{dy}{dx}$
$xy$	$1 \cdot y + x\dfrac{dy}{dx}$ (product rule)

Terms in $x$ alone are differentiated as usual. Every time you differentiate something with $y$ in it, a $\dfrac{dy}{dx}$ appears.

The steps

1. Differentiate both sides with respect to $x$.
2. Move all terms containing $\dfrac{dy}{dx}$ to one side, and everything else to the other.
3. Factor out $\dfrac{dy}{dx}$.
4. Divide to solve for $\dfrac{dy}{dx}$.

The answer usually contains both $x$ and $y$. That’s expected: to find a slope, you need the whole point $(x, y)$, not just $x$. (On a circle, two points share each $x$-value, with different slopes.)

Tangent lines, horizontal and vertical

To find the slope at a point, substitute both coordinates into $\dfrac{dy}{dx}$. Then write the tangent line in point-slope form, $y - y_1 = m(x - x_1)$.

If $\dfrac{dy}{dx} = \dfrac{N}{D}$:

• Horizontal tangent: $N = 0$ and $D \ne 0$.
• Vertical tangent: $D = 0$ and $N \ne 0$.

Either way, the point must also be on the curve, so solve together with the original equation. You’ll analyze implicit curves more in implicit relations analysis.

Worked examples

Example 1: A circle

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

Solution. Differentiate both sides:

$$2x + 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{x}{y}$$

At $(3, 4)$, the slope is $-\dfrac{3}{4}$, so the tangent line is

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

Check: the radius from $(0, 0)$ to $(3, 4)$ has slope $\dfrac{4}{3}$, and a tangent to a circle is perpendicular to the radius. Indeed, $-\dfrac{3}{4}$ is the negative reciprocal of $\dfrac{4}{3}$.

The tangent to $x^2 + y^2 = 25$ at $(3, 4)$ has slope $-\tfrac{3}{4}$, found implicitly.

Example 2: A product term

Find $\dfrac{dy}{dx}$ for $x^2 + xy + y^2 = 7$, then find the slope at $(1, 2)$.

Solution. Use the product rule on $xy$:

$$2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$$

Collect the $\dfrac{dy}{dx}$ terms and factor:

$$\begin{aligned} (x + 2y)\frac{dy}{dx} &= -2x - y \\ \frac{dy}{dx} &= -\frac{2x + y}{x + 2y} \end{aligned}$$

First check that $(1, 2)$ is on the curve: $1 + 2 + 4 = 7$. Then

$$\frac{dy}{dx} = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5}$$

Example 3: A trig term

Find the slope of the curve $x\cos y + y = 1$ at the point $(1, 0)$.

Solution. Product rule on $x\cos y$, with the chain rule on $\cos y$:

$$\cos y + x\left(-\sin y\frac{dy}{dx}\right) + \frac{dy}{dx} = 0$$ $$\frac{dy}{dx}(1 - x\sin y) = -\cos y \quad\Rightarrow\quad \frac{dy}{dx} = \frac{-\cos y}{1 - x\sin y}$$

At $(1, 0)$ (check: $1 \cdot \cos 0 + 0 = 1$), the slope is $\dfrac{-1}{1 - 0} = -1$.

Example 4: Horizontal and vertical tangents

For the curve $x^2 - xy + y^2 = 3$, find all points with a horizontal tangent and all points with a vertical tangent.

Solution. Differentiate:

$$2x - \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{y - 2x}{2y - x}$$

Horizontal: set the numerator to zero, so $y = 2x$. Substitute into the curve:

$$x^2 - 2x^2 + 4x^2 = 3 \quad\Rightarrow\quad 3x^2 = 3 \quad\Rightarrow\quad x = \pm 1$$

The points are $(1, 2)$ and $(-1, -2)$. The denominator there is $2(2) - 1 = 3$ and $2(-2) + 1 = -3$, not zero, so these tangents really are horizontal.

Vertical: set the denominator to zero, so $x = 2y$. Substitute:

$$4y^2 - 2y^2 + y^2 = 3 \quad\Rightarrow\quad y = \pm 1$$

The points are $(2, 1)$ and $(-2, -1)$, where the numerator is $1 - 4 = -3$ and $-1 + 4 = 3$, not zero.

Common mistakes

Forgetting the dy/dx on y terms. $\dfrac{d}{dx}(y^2)$ is $2y\dfrac{dy}{dx}$, not $2y$. If your final answer has no $\dfrac{dy}{dx}$ to solve for, you’ve dropped it somewhere.

Skipping the product rule on xy. $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$. Writing just $x\dfrac{dy}{dx}$, or just $1 \cdot \dfrac{dy}{dx}$, is one of the most common AP errors.

Not differentiating the right side. Differentiate both sides. A constant like $25$ becomes $0$, and a term like $3x$ on the right becomes $3$.

Messy algebra at the end. Get every $\dfrac{dy}{dx}$ term on one side and factor it out before dividing. Don’t divide by only one of the terms.

Using a point that isn’t on the curve. Before substituting, check that the point satisfies the original equation. For horizontal or vertical tangents, you must solve together with the original equation to find actual points.

Forgetting to check the other part of the fraction. A horizontal tangent needs the numerator $0$ and the denominator not $0$. If both are $0$, the slope isn’t determined by this formula, and you’d need more work.

Practice

1. (Warm-up) Find $\dfrac{d}{dx}$ of each expression, assuming $y$ is a function of $x$.

• (a) $y^4$
• (b) $xy$
• (c) $\sin y$

Solution

(a) $4y^3\dfrac{dy}{dx}$

(b) $y + x\dfrac{dy}{dx}$

(c) $\cos y\dfrac{dy}{dx}$

2. (Warm-up) Find the slope of the circle $x^2 + y^2 = 100$ at the point $(6, -8)$.

Solution

As in Example 1, $\dfrac{dy}{dx} = -\dfrac{x}{y}$. At $(6, -8)$:

$$\frac{dy}{dx} = -\frac{6}{-8} = \frac{3}{4}$$

3. (Warm-up) For the parabola $y^2 = 4x$, find $\dfrac{dy}{dx}$ and the slope at $(1, 2)$.

Solution

$$2y\frac{dy}{dx} = 4 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2}{y}$$

At $(1, 2)$, the slope is $\dfrac{2}{2} = 1$.

4. (Core) Find the equation of the tangent line to $x^3 + y^3 = 9$ at $(1, 2)$.

Solution

Check: $1 + 8 = 9$. Differentiate:

$$3x^2 + 3y^2\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{x^2}{y^2}$$

At $(1, 2)$, the slope is $-\dfrac{1}{4}$, so the tangent line is

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

5. (Core) Find $\dfrac{dy}{dx}$ for $xy + y^2 = 6$, and evaluate it at $(1, 2)$.

Solution

$$y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{y}{x + 2y}$$

Check $(1, 2)$: $2 + 4 = 6$. Then $\dfrac{dy}{dx} = -\dfrac{2}{1 + 4} = -\dfrac{2}{5}$.

6. (Core) Find $\dfrac{dy}{dx}$ for $y + \ln y = x$, and the slope at $(1, 1)$.

Solution

$$\frac{dy}{dx} + \frac{1}{y}\frac{dy}{dx} = 1 \quad\Rightarrow\quad \frac{dy}{dx}\left(\frac{y + 1}{y}\right) = 1 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{y}{y + 1}$$

Check $(1, 1)$: $1 + \ln 1 = 1$. The slope is $\dfrac{1}{2}$.

7. (Core) Consider the ellipse $x^2 + 4y^2 = 8$.

• (a) Find the equation of the tangent line at $(2, 1)$.
• (b) Find the points where the tangent line is horizontal.

Solution

$$2x + 8y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{x}{4y}$$

(a) Check: $4 + 4 = 8$. The slope at $(2, 1)$ is $-\dfrac{2}{4} = -\dfrac{1}{2}$, so the tangent line is $y - 1 = -\dfrac{1}{2}(x - 2)$, or $y = -\dfrac{1}{2}x + 2$.

(b) Horizontal when $x = 0$ (and $y \ne 0$). Then $4y^2 = 8$, so $y = \pm\sqrt{2}$. The points are $(0, \sqrt{2})$ and $(0, -\sqrt{2})$.

8. (Challenge) Find all points on $x^2 + xy + y^2 = 12$ where the tangent line is horizontal.

Solution

As in Example 2, $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$.

Horizontal when $2x + y = 0$, so $y = -2x$. Substitute:

$$x^2 + x(-2x) + (-2x)^2 = 12 \quad\Rightarrow\quad 3x^2 = 12 \quad\Rightarrow\quad x = \pm 2$$

The points are $(2, -4)$ and $(-2, 4)$. The denominators are $2 - 8 = -6$ and $-2 + 8 = 6$, both nonzero, so both tangents are horizontal.

9. (Challenge) Consider the curve $e^y = x^2 + y$.

• (a) Show that $(1, 0)$ is on the curve.
• (b) Find $\dfrac{dy}{dx}$.
• (c) What happens to the tangent line at $(1, 0)$?

Solution

(a) $e^0 = 1$ and $1^2 + 0 = 1$, so yes.

(b)

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

(c) At $(1, 0)$, the denominator is $e^0 - 1 = 0$ while the numerator is $2 \ne 0$. So the slope is undefined and the tangent line is vertical: it’s the line $x = 1$.