Implicit Differentiation Calculator

About the Implicit Differentiation Calculator

Many curves cannot be written neatly as $y = f(x)$ . A circle such as $x^2 + y^2 = 25$ fails the vertical line test, and a curve like the folium of Descartes, $x^3 + y^3 = 6xy$ , has no convenient formula for $y$ at all. Their slopes still exist at almost every point, and implicit differentiation is the technique that finds them without solving for $y$ first.

This calculator performs the method exactly as it is taught. It treats $y$ as an unknown function of $x$ , differentiates both sides of your equation with respect to $x$ , and attaches a factor of $y'$ to every term containing $y$ through the chain rule. It then gathers the $y'$ terms on one side, factors $y'$ out and divides, giving a formula for $\frac{dy}{dx}$ in terms of both $x$ and $y$ .

If you supply a point on the curve, the calculator substitutes it to give the numerical slope and writes the equation of the tangent line there. The final formula is verified independently by comparing it with $-\frac{F_x}{F_y}$ , the result predicted by partial derivatives, so you can trust the answer before you copy it into your work.

What is implicit differentiation?

An equation such as $x^2 + y^2 = 25$ defines $y$ implicitly: it describes a relationship between $x$ and $y$ rather than giving $y$ directly. Near most points of the curve, $y$ still behaves like a function of $x$ , even if you cannot write that function down. Implicit differentiation finds its derivative from the relationship itself.

The key fact is the chain rule. If $y$ depends on $x$ , then the derivative of $y^2$ with respect to $x$ is not $2y$ but $2y \cdot \frac{dy}{dx}$ . The same goes for $\sin y$ , $e^y$ or any other expression in $y$ : differentiate as usual, then multiply by $y'$ . Terms that contain only $x$ are differentiated normally, and constants vanish.

The result usually contains both $x$ and $y$ , as in $\frac{dy}{dx} = -\frac{x}{y}$ . That is expected and correct. A single $x$ -value can correspond to several points on the curve, and the slope is different at each of them, so the formula needs the $y$ -coordinate to know which point you mean.

How to use the Implicit Differentiation Calculator

Type an equation containing x and y with one equals sign, for example x^2 + y^2 = 25 or x^3 + y^3 = 6 x y.
Use a space or * for multiplication between letters, so x y or x*y for the product of x and y.
Optionally enter a point on the curve, such as x = 3, y = 4, to get the slope and tangent line there.
Press Find dy/dx to see each side differentiated, the y′ terms collected and the final formula.
If you gave a point, read off the slope and the tangent line equation from the last steps.

Formulas

Chain rule for a y term

\frac{d}{dx}\big[g(y)\big] = g'(y)\,\frac{dy}{dx}

Product rule with y

\frac{d}{dx}\big[xy\big] = y + x\,\frac{dy}{dx}

Shortcut from partial derivatives

F(x,y) = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{F_x}{F_y}

Tangent line at (x₀, y₀)

y - y_0 = m\,(x - x_0),\quad m = \left.\frac{dy}{dx}\right|_{(x_0,\,y_0)}

Worked example: x² + y² = 25 at (3, 4)

The worked example below is computed live for the circle $x^2 + y^2 = 25$ at the point $(3, 4)$ . Differentiating both sides gives $2x + 2y\,y' = 0$ ; the $2y\,y'$ term comes from the chain rule. Moving $2x$ across and dividing by $2y$ gives $\frac{dy}{dx} = -\frac{x}{y}$ .

At $(3, 4)$ the slope is $-\frac34 = -0.75$ , so the tangent line is $y - 4 = -0.75(x - 3)$ . Geometry confirms it: the radius to $(3, 4)$ has slope $\frac43$ , and a tangent to a circle is perpendicular to the radius, so its slope must be the negative reciprocal, $-\frac34$ .

The solution below is generated by the same engine that powers the calculator above.

Final answer

\frac{dy}{dx} = -\frac{x}{y},\quad \text{slope at the point} = -0.75

How we read your problem

Find dy/dx for a curve defined implicitly, treating y as a differentiable function of x.

{x}^{2}+{y}^{2} = 25

Step-by-step solution

Treat y as a function of x and differentiate both sides with respect to x
Every term containing y needs the chain rule: d/dx[g(y)] = g′(y)·y′.
$\frac{d}{dx}\left[{x}^{2}+{y}^{2}\right] = \frac{d}{dx}\left[25\right]$
Result of differentiating each side
$2x + \left(2y\right) y' = 0$
Collect the y′ terms on one side
Move every y′ term to the left and everything else to the right, then factor out y′.
$\left(2y\right) y' = -\left(2x\right)$
Divide to isolate y′
$\frac{dy}{dx} = -\frac{x}{y}$
Substitute x = 3, y = 4
$\left.\frac{dy}{dx}\right|_{(3,\,4)} = -0.75$
Tangent line at the point
$y - 4 = -0.75\,(x - 3)$

Formulas used

Chain rule for y
$\frac{d}{dx}\,g(y) = g'(y)\,\frac{dy}{dx}$
Implicit function formula
$\frac{dy}{dx} = -\frac{F_x}{F_y},\quad F(x,y) = 0$

Differentiating both sides keeps the equation balanced; because y depends on x, each y-term produces a factor y′, which is then solved for algebraically. The result usually contains both x and y.

Verification

Matches −F_x / F_y from an independent symbolic differentiator at several sample points ✓
(3, 4) lies on the curve ✓

Detailed explanation

The method in four steps

First, differentiate every term on both sides with respect to $x$ . Second, wherever a term contains $y$ , multiply its derivative by $y'$ ; wherever $x$ and $y$ are multiplied, use the product rule. Third, move every term containing $y'$ to one side and everything else to the other. Fourth, factor out $y'$ and divide by what remains.

Applied to $x^3 + y^3 = 6xy$ , the first two steps give $3x^2 + 3y^2 y' = 6y + 6x y'$ , because the right side is a product. Collecting terms gives $(3y^2 - 6x)\,y' = 6y - 3x^2$ , so $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$ . The calculator shows exactly this sequence.

Why the chain rule produces y′

Think of $y$ as $y(x)$ , a function you cannot see. Then $y^3$ is really a composite function, the cube of $y(x)$ , and the chain rule says its derivative is $3y(x)^2 \cdot y'(x)$ . Implicit differentiation is nothing more than the ordinary chain rule applied to an inner function whose formula is unknown.

This is also why $\frac{d}{dx}[\sin y] = \cos y \cdot y'$ and $\frac{d}{dx}[e^y] = e^y\,y'$ . For the equation $\sin y = x$ , differentiating gives $\cos y\,y' = 1$ , so $y' = \frac{1}{\cos y}$ . Since this curve is $y = \arcsin x$ on its principal branch, the result is a quick derivation of the derivative of arcsine once $\cos y$ is written as $\sqrt{1 - x^2}$ .

Slopes, tangent lines and special points

Substituting a point into the formula gives the slope there, but only if the point is actually on the curve; it is worth checking that it satisfies the original equation first. With the slope $m$ and the point $(x_0, y_0)$ , the tangent line is $y - y_0 = m(x - x_0)$ .

The formula also locates special features. Horizontal tangents occur where the numerator of $\frac{dy}{dx}$ is zero and the denominator is not. Vertical tangents occur where the denominator is zero and the numerator is not; on the circle, that happens at $(\pm 5, 0)$ , where $y = 0$ . If both numerator and denominator vanish, as at the origin on the folium, the curve may cross itself and the slope must be investigated separately.

A shortcut with partial derivatives

Moving everything to one side writes the curve as $F(x, y) = 0$ . Multivariable calculus then gives the formula $\frac{dy}{dx} = -\frac{F_x}{F_y}$ , where $F_x$ and $F_y$ are partial derivatives. For $x^2 + xy + y^2 = 7$ , $F_x = 2x + y$ and $F_y = x + 2y$ , so $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$ , which is $-0.8$ at $(1, 2)$ .

The calculator uses this formula as an independent check on the step-by-step result. You can use it the same way: when your hand calculation and the shortcut agree, you can be confident in the answer.

Where the technique is used next: related rates

Implicit differentiation is the engine behind related-rates problems. When a ladder slides down a wall, its foot and top satisfy $x^2 + y^2 = L^2$ at every moment, and both coordinates depend on time. Differentiating with respect to $t$ instead of $x$ gives $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ , which connects the speed of the foot to the speed of the top.

The pattern is identical to the one on this page: differentiate the relationship, apply the chain rule to every quantity that changes, and solve for the rate you want. It also underlies logarithmic differentiation, where you take $\ln$ of both sides of $y = x^x$ and differentiate $\ln y = x \ln x$ implicitly to get $y' = x^x(\ln x + 1)$ .

Common mistakes

Writing the derivative of $y^2$ as $2y$ and forgetting the factor $y'$ .
Missing the product rule on mixed terms, so that $\frac{d}{dx}[xy]$ is written as $y'$ or $xy'$ alone.
Differentiating the constant on the right side as if it were a variable instead of giving 0.
Dividing by an expression that is zero at the point of interest, which hides a vertical tangent.
Substituting a point that does not satisfy the original equation.
Solving for $y$ first when it is not necessary, which often creates square roots with sign ambiguities.

Tips

Circle or highlight every $y'$ as you write it so none get lost when collecting terms.
Expand products before differentiating if it reduces the number of product-rule terms.
Use $-\frac{F_x}{F_y}$ to check your result quickly.
Simplify the final fraction by cancelling common factors such as 2 or 3.
Verify that your point lies on the curve before finding the tangent line.
For a second derivative, differentiate $y'$ implicitly again and substitute the first-derivative formula.

Frequently asked questions

When should I use implicit differentiation?

Whenever y is tied to x by an equation that is hard or impossible to solve for y, such as x³ + y³ = 6xy, or when solving would split the curve into several branches, as with circles and ellipses.

Why does dy/dx contain y as well as x?

One x-value can correspond to several points on an implicit curve, each with its own slope. The y in the formula identifies which point you mean. For the circle, −x/y gives −0.75 at (3, 4) and +0.75 at (3, −4).

How do I enter the point for the slope?

Type both coordinates in the optional box, for example x = 3, y = 4. The calculator substitutes them into dy/dx and writes the tangent line in point-slope form.

Can it differentiate equations with sin y or e^y?

Yes. Any differentiable expression in x and y can be used. Terms like sin y and e^y receive the factor y′ through the chain rule, so sin y = x gives dy/dx = 1/cos y.

What does it mean if the denominator of dy/dx is zero?

At such a point the tangent line is vertical, provided the numerator is not also zero. On x² + y² = 25 this happens at (5, 0) and (−5, 0).

Is implicit differentiation the same as using partial derivatives?

They give the same answer. The step-by-step method uses the chain rule directly, while the formula dy/dx = −F_x/F_y uses partial derivatives of F(x, y). The calculator shows the first and checks with the second.

Can I find d²y/dx² with this calculator?

The calculator returns the first derivative. For the second, differentiate the formula for dy/dx again with the quotient rule, remembering that y still depends on x, and replace each y′ that appears with the first-derivative expression.

Which letters must the equation use?

The equation must be written in x and y, with y treated as the dependent variable. Other letters are not supported, and the calculator says so rather than guessing their role.

Do I need to solve for y before differentiating?

No, and it is usually better not to. Solving x² + y² = 25 for y gives ±√(25 − x²), two separate functions with awkward square roots. Differentiating the original equation handles both halves of the circle at once.

Differentiate your implicit equation

Enter an equation in x and y above to find dy/dx with every chain-rule step, and add a point to get the slope and tangent line.

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