Implicit differentiation finds \( \frac{dy}{dx} \) when an equation mixes \( x \) and \( y \) and you can’t (or don’t want to) solve for \( y \) first, as in \( x^2 + y^2 = 25 \) or \( x^3 + y^3 = 6xy \).
The key idea: treat \( y \) as an unknown function of \( x \). So whenever you differentiate something containing \( y \), the chain rule adds a factor of \( \frac{dy}{dx} \):
$$\frac{d}{dx}y^2 = 2y\,\frac{dy}{dx}, \qquad \frac{d}{dx}\sin y = \cos y\,\frac{dy}{dx}$$
Explicit vs. implicit equations
An equation like \( y = x^3 - 4x \) is explicit: \( y \) is already isolated, so you differentiate the right side and you are done. An equation like \( x^2 + y^2 = 25 \) is implicit: it describes a curve, but \( y \) is tangled up with \( x \).
Sometimes you could solve for \( y \), but the result is messy or splits into several pieces (the circle needs both \( +\sqrt{\ } \) and \( -\sqrt{\ } \)). Sometimes you can’t solve for \( y \) at all with ordinary algebra, as with \( x^3 + y^3 = 6xy \) or \( \sin(xy) = x \). Implicit differentiation handles every one of these the same way, without ever isolating \( y \).
Why it works
Near most points, a curve like a circle looks like the graph of some function \( y(x) \), even if you never write that function down. If two expressions are equal for every \( x \) along the curve, then their rates of change are equal too. So you can differentiate both sides and keep the equation true.
The only subtlety is that \( y \) changes when \( x \) changes. Differentiating \( y^2 \) with respect to \( x \) is exactly like differentiating \( (\text{inner function})^2 \): the outer derivative is \( 2y \), and the chain rule multiplies by the inner derivative, \( \frac{dy}{dx} \). Terms that contain only \( x \) are differentiated as usual, with no extra factor.
The 4-step method
- Differentiate both sides of the equation with respect to \( x \).
- Every time you differentiate a \( y \) term, multiply by \( \frac{dy}{dx} \) (often written \( y' \)).
- Collect all \( y' \) terms on one side and everything else on the other.
- Factor out \( y' \) and divide.
Example 1: a circle
\( x^2 + y^2 = 25 \). Differentiate:
$$2x + 2y\,y' = 0 \quad\Longrightarrow\quad y' = -\frac{x}{y}$$
At the point \( (3, 4) \), the slope is \( -\frac34 \). Check: the top half of the circle is \( y = \sqrt{25 - x^2} \), whose ordinary derivative is \( \frac{-x}{\sqrt{25 - x^2}} = -\frac{x}{y} \). Same answer, but implicit differentiation got there without square roots.
The formula also makes geometric sense. The radius to \( (x, y) \) has slope \( \frac{y}{x} \), and \( -\frac{x}{y} \) is its negative reciprocal, so the tangent line is perpendicular to the radius, just as you learned in geometry.
Example 2: the folium of Descartes
\( x^3 + y^3 = 6xy \). The right side needs the product rule:
$$3x^2 + 3y^2y' = 6y + 6x\,y'$$
Collect the \( y' \) terms:
$$3y^2y' - 6xy' = 6y - 3x^2 \quad\Longrightarrow\quad y' = \frac{2y - x^2}{y^2 - 2x}$$
At \( (3, 3) \), which is on the curve since \( 27 + 27 = 54 \): \( y' = \frac{6 - 9}{9 - 6} = -1 \).
Example 3: trig and exponentials
\( \sin(xy) = x \). Chain rule on the left, product rule inside it:
$$\cos(xy)\,(y + x\,y') = 1 \quad\Longrightarrow\quad y' = \frac{1 - y\cos(xy)}{x\cos(xy)}$$
The expression looks complicated, but each piece came from one rule applied in order: chain rule for the outer sine, product rule for \( xy \), then algebra to isolate \( y' \).
Example 4: an exam-style tangent line
Find the tangent line to \( x^2y + y^3 = 10 \) at \( (3, 1) \). First confirm the point is on the curve: \( 9 + 1 = 10 \). Now differentiate. The first term is a product, so it gives two terms:
$$2xy + x^2y' + 3y^2y' = 0$$
Move \( 2xy \) across and factor:
$$y' = -\frac{2xy}{x^2 + 3y^2}$$
At \( (3, 1) \) the slope is \( -\frac{6}{9 + 3} = -\frac12 \), so the tangent line is \( y - 1 = -\frac12(x - 3) \). A good habit: plug the point in right after differentiating, before simplifying. The numbers are often easier than the symbols.
Example 5: horizontal and vertical tangents
Where does the circle \( x^2 + y^2 = 25 \) have horizontal or vertical tangent lines? Use \( y' = -\frac{x}{y} \).
- Horizontal tangents need \( y' = 0 \), so the numerator is zero: \( x = 0 \). The curve then gives \( y = \pm 5 \), so the points are \( (0, 5) \) and \( (0, -5) \).
- Vertical tangents happen where the denominator is zero and the numerator isn’t: \( y = 0 \), giving \( (5, 0) \) and \( (-5, 0) \).
The same “numerator zero, denominator zero” test works on any implicit curve, and it is a common exam question.
Finding a tangent line
Once you have \( y' \) at a point, the tangent line is \( y - y_0 = m(x - x_0) \). For the circle at \( (3,4) \): \( y - 4 = -\frac34(x - 3) \).
Second derivatives implicitly
For the circle, differentiate \( y' = -\frac{x}{y} \) with the quotient rule and substitute \( y' \) back:
$$\begin{aligned} y'' &= -\frac{y - x y'}{y^2} = -\frac{y + x^2/y}{y^2} \\ &= -\frac{x^2 + y^2}{y^3} = -\frac{25}{y^3} \end{aligned}$$
On the top half, \( y > 0 \), so \( y'' < 0 \) and the curve is concave down, which matches the picture of an upside-down bowl. The last step used the original equation \( x^2 + y^2 = 25 \) to simplify; watch for that move, because it often turns a messy answer into a clean one.
Proving inverse function derivatives
Implicit differentiation is how the standard derivative formulas for inverse functions are proved.
Natural log. If \( y = \ln x \), then \( e^y = x \). Differentiate both sides: \( e^y\,y' = 1 \), so \( y' = \frac{1}{e^y} = \frac1x \).
Arcsine. If \( y = \arcsin x \), then \( \sin y = x \). Differentiate: \( \cos y\,y' = 1 \), so \( y' = \frac{1}{\cos y} \). Since \( \cos y = \sqrt{1 - \sin^2y} = \sqrt{1 - x^2} \) on the range of arcsine,
$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$$
Common mistakes
- Forgetting the \( y' \). Writing \( \frac{d}{dx}y^3 = 3y^2 \) is the most common error. It must be \( 3y^2y' \).
- Missing the product rule. A term like \( xy \) or \( x^2y \) is a product of two functions of \( x \). Its derivative has two terms, for example \( y + xy' \).
- Differentiating the constant wrong. The right side of \( x^2 + y^2 = 25 \) becomes 0, not 25.
- Stopping too early. If \( y' \) still appears on both sides, you haven’t finished: collect those terms and factor.
- Using a point that isn’t on the curve. Always substitute the point into the original equation first. A slope at a point off the curve means nothing.
Where it’s used
- Inverse function derivatives. Proving the derivatives of ln x, arctan x and arcsin x, as shown above.
- Related rates. In related rates problems every variable depends on time, so you differentiate with respect to \( t \) instead of \( x \). The mechanics are identical: a ladder whose base and top satisfy \( x^2 + y^2 = 100 \) gives \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \), which links the speed of the top to the speed of the base.
- Logarithmic differentiation. For functions like \( x^x \), you take the log of both sides and then differentiate implicitly. See logarithmic differentiation.
- Curves in geometry and physics. Ellipses, hyperbolas, level curves of temperature or pressure and orbits are all naturally described by equations rather than formulas for \( y \), and their slopes come from implicit differentiation.
Check an explicit branch with the solver
When you can solve for \( y \), the solver confirms your implicit answer. Here is the top half of the circle:
The derivative calculator works the same way for any explicit branch you type in.
Practice problems
Try these before checking the answers.
- \( x^2 + xy + y^2 = 7:\ \text{find } y' \)
- Slope of \( x^2 + xy + y^2 = 7 \) at \( (1, 2) \)
- \( y^2 = x^3:\ \text{find } y' \)
- \( xy = 1:\ \text{find } y' \)
- Slope of \( xy + y^2 = 6 \) at \( (1, 2) \)
- Points where \( x^2 + y^2 = 25 \) has a vertical tangent
Answers: (1) \( y' = -\frac{2x + y}{x + 2y} \); (2) \( -\frac45 \); (3) \( y' = \frac{3x^2}{2y} \); (4) \( y' = -\frac{y}{x} \), which equals \( -\frac{1}{x^2} \) since \( y = \frac1x \); (5) \( -\frac25 \), from \( y' = -\frac{y}{x + 2y} \); (6) \( (5, 0) \) and \( (-5, 0) \).
FAQ
Why do we multiply by dy/dx?
Because \( y \) depends on \( x \). Differentiating \( y^2 \) with respect to \( x \) is a chain rule problem with \( y \) as the inside function.
Can dy/dx contain y?
Yes. Implicit derivatives usually depend on both coordinates, which is why you need a specific point to get a number.
When should I use implicit differentiation?
Use it whenever \( y \) is hard or impossible to isolate, when solving for \( y \) would create several branches, or when a problem gives you a point and asks for a slope on a curve defined by an equation.
Is implicit differentiation the same as explicit differentiation?
It gives the same derivative whenever both methods apply, as the circle example shows. The difference is only in the route: explicit differentiation needs \( y \) isolated first, implicit differentiation doesn’t.
Do I need a point to finish the problem?
Only if you want a number. The general formula for \( y' \) in terms of \( x \) and \( y \) is a complete answer; plug in a point on the curve to get a specific slope.
Further reading
- Implicit Differentiation (Paul’s Online Math Notes) — many more worked examples, including exponentials and logarithms.
- 3.8 Implicit Differentiation (OpenStax Calculus Volume 1) — a textbook treatment with a problem-solving strategy and exercises.
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