Implicit Differentiation, Step by Step

Explicit and implicit equations

An explicit equation gives $y$ directly as a formula in $x$, such as $y = x^2 + 1$. Many important curves are instead described implicitly, by an equation linking $x$ and $y$. The circle $x^2 + y^2 = 25$ is a familiar example. Solving it for $y$ gives two functions, $y = \sqrt{25 - x^2}$ for the top half and $y = -\sqrt{25 - x^2}$ for the bottom, and differentiating each separately is awkward.

Other curves cannot be solved for $y$ at all with ordinary algebra, such as $x^3 + y^3 = 6xy$ or $e^y + xy = e$. Their graphs still have tangent lines at almost every point, and implicit differentiation is the tool that finds the slopes. The method relies entirely on the chain rule, so it is worth reviewing derivative rules explained first if that rule feels shaky.

The method in four steps

1. Differentiate both sides of the equation with respect to $x$.
2. Whenever you differentiate an expression containing $y$, multiply by $\frac{dy}{dx}$, because $y$ is a function of $x$ and the chain rule applies.
3. Collect every term containing $\frac{dy}{dx}$ on one side and everything else on the other.
4. Factor out $\frac{dy}{dx}$ and divide to isolate it.

The key fact behind step 2 is this chain rule identity, where $f$ is any differentiable function:

So $\frac{d}{dx}y^2 = 2y\frac{dy}{dx}$, $\frac{d}{dx}\sin y = \cos y\,\frac{dy}{dx}$ and $\frac{d}{dx}e^y = e^y\frac{dy}{dx}$. Products such as $xy$ need the product rule as well: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$.

Example: the circle

Notice that the implicit answer works for both halves at once: at $(3, -4)$ the slope is $+\frac{3}{4}$. The formula also shows where the tangents are horizontal, when $x = 0$, at $(0, 5)$ and $(0, -5)$, and where they are vertical, when $y = 0$, at $(5, 0)$ and $(-5, 0)$. The answer contains $y$ as well as $x$, which is normal for implicit derivatives; you need a point on the curve, not just an $x$ value, to get a number.

Examples that cannot be solved for y

Second derivatives

To find $\frac{d^2y}{dx^2}$, differentiate the first derivative implicitly again, then substitute the first derivative back in. For the circle, start from $y' = -\frac{x}{y}$ and use the quotient rule:

The last step uses the original equation, $x^2 + y^2 = 25$, to simplify, a trick that often helps. The result is negative on the top half, where $y > 0$, so the curve is concave down there, and positive on the bottom half, which matches the picture of a circle.

Deriving inverse function derivatives

Implicit differentiation explains where the derivatives of inverse functions come from. Let $y = \arctan x$. Then $\tan y = x$. Differentiating gives $\sec^2 y\,\frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$.

The same approach with $y = \ln x$, rewritten as $e^y = x$, gives $e^y\frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$. In general, the slope of an inverse function is the reciprocal of the original slope at the matching point, as described in our article on inverse functions.

Related rates

When the variables depend on time, the same idea works with respect to $t$ instead of $x$. Every variable gets its own rate.

Common mistakes

• Forgetting the $\frac{dy}{dx}$ factor when differentiating a term with $y$.
• Treating $xy$ as a single variable instead of using the product rule.
• Differentiating the constant on the right as if it were a variable; constants give 0.
• Substituting numbers before differentiating, which freezes quantities that are changing.
• Using a point that is not actually on the curve. Always check the point first.

The implicit differentiation calculator carries out every step, collects the derivative terms and can evaluate the slope at a point. For explicit functions, use the derivative calculator.

Why the method is valid

Implicit differentiation assumes that near the point of interest, the curve really is the graph of some differentiable function $y(x)$, even if we cannot write that function down. Substituting it into the equation turns the equation into an identity in $x$, true for every $x$ near the point. Two expressions that are equal for every $x$ must have equal derivatives, which is why we may differentiate both sides.

The assumption fails only at special points, such as where the tangent is vertical or where the curve crosses itself. The folium of Descartes crosses itself at the origin, and there the derivative formula gives $\frac{0}{0}$. At such points the curve is not locally the graph of a single function, so it has no single slope. The implicit function theorem, met in multivariable calculus, states exactly when the assumption is safe.

A shortcut with partial derivatives

If the curve is written as $F(x, y) = 0$, there is a compact formula for the slope that follows from the same reasoning:

Here $F_x$ and $F_y$ are partial derivatives, found by differentiating with respect to one variable while treating the other as a constant. For the circle, $F = x^2 + y^2 - 25$ gives $F_x = 2x$ and $F_y = 2y$, so the slope is $-\frac{x}{y}$, as before. The partial derivative calculator computes $F_x$ and $F_y$ for you.

Practice problems with answers

1. $x^2 - y^2 = 9$. Answer: $\frac{dy}{dx} = \frac{x}{y}$.
2. $xy = 12$, slope at $(3, 4)$. Answer: $\frac{dy}{dx} = -\frac{y}{x}$, which is $-\frac{4}{3}$ there.
3. $y^3 + y = x$. Answer: $\frac{dy}{dx} = \frac{1}{3y^2 + 1}$, which is always positive, so the curve always rises.
4. $\cos y = x$. Answer: $\frac{dy}{dx} = -\frac{1}{\sin y}$, which leads to the derivative of arccosine.

For each one, check your answer by choosing a point on the curve, estimating the slope numerically from a nearby point, and comparing.

Where implicit curves appear

Implicit equations are common in practice. Conic sections such as ellipses and hyperbolas are naturally written implicitly. In economics, an indifference curve links quantities of two goods that give a consumer the same satisfaction, and its slope, found implicitly, is the rate at which the consumer will trade one good for the other. In thermodynamics, equations of state link pressure, volume and temperature, and implicit differentiation gives how one changes when another is varied. Whenever two quantities are tied together by a rule rather than a formula, this method gives their rate of change.

Frequently asked questions