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Implicit differentiation calculator

Differentiate an implicit equation

Enter one equation relating x and y, then solve the differentiated relation for dy over dx without losing dependent-variable factors.

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Try an example

Result

Implicit derivative

dydx=−xy\frac{dy}{dx}=-\frac{x}{y}
Conditions
  • y is treated as a differentiable function of x along the curve.
  • The displayed slope formula requires y not equal to zero.

Steps

  1. Differentiate both sides. Apply the derivative with respect to x to every term.2x+2ydydx=02x+2y\frac{dy}{dx}=0
  2. Collect derivative terms. Move the term without dy over dx to the opposite side.2ydydx=−2x2y\frac{dy}{dx}=-2x
  3. Solve for the derivative. Divide by 2y and keep the nonzero-denominator condition.dydx=−xy,y≠0\frac{dy}{dx}=-\frac{x}{y},\qquad y\ne0
Independent check

Substitution into the differentiated equation gives 2x + 2y(-x/y) = 0 wherever y is nonzero.

Scope

What this implicit differentiation calculator covers

Implicit differentiation finds dy over dx when an equation defines y in terms of x without first isolating y. Differentiate both sides with respect to x, attach dy over dx to each derivative involving y, and solve the resulting linear equation for the derivative.

Polynomial relations

Differentiate equations containing powers and products of x and y without isolating y first.

Examples: circles, ellipses, algebraic curves

Products containing x and y

Apply the product rule when both factors vary with x, including the term xy.

Examples: xy plus y squared, x cubed plus y cubed equals 6xy

Composite y-expressions

Attach dy over dx through the chain rule when a trigonometric, exponential, or logarithmic function contains y.

Examples: sine of x plus y, natural logarithm of y

Slope restrictions

Keep the denominator that results from solving for dy over dx and identify points where the displayed finite-slope formula does not apply.

Examples: vertical tangents, singular points

How to use it

Enter enough information for one clear task

  1. 1
    Enter an equation

    Include one equals sign and use parentheses to make powers, products, and function arguments unambiguous.

  2. 2
    Confirm x and y roles

    This page treats x as independent and y as a differentiable function of x along the equation's curve.

  3. 3
    Follow every chain and product rule

    Each derivative involving y needs a dy over dx factor, and an xy term needs the full product rule.

  4. 4
    Inspect the denominator

    The final formula is valid only where its denominator is nonzero and the original equation and functions are defined.

Worked inputs

Examples to try

Use these examples to recognize the method, compare equivalent forms, and check your own work.

Circle

chain rule on y squared

x2+y2=25x^2+y^2=25

Expected result

dydx=−xy\frac{dy}{dx}=-\frac{x}{y}

Cubic relation

chain and product rules

x3+y3=6xyx^3+y^3=6xy

Expected result

dydx=2y−x2y2−2x\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}

Product relation

product rule

xy+y2=4xy+y^2=4

Expected result

dydx=−yx+2y\frac{dy}{dx}=-\frac{y}{x+2y}

Mixed quadratic

product and chain rules

x2+xy−y2=7x^2+xy-y^2=7

Expected result

dydx=2x+y2y−x\frac{dy}{dx}=\frac{2x+y}{2y-x}

Trigonometric relation

chain rule

sin⁡(x+y)=x\sin(x+y)=x

Expected result

dydx=sec⁡(x+y)−1\frac{dy}{dx}=\sec(x+y)-1

Logarithmic relation

logarithmic chain and product rules

ln⁡y+xy=2\ln y+xy=2

Expected result

dydx=−y21+xy\frac{dy}{dx}=-\frac{y^2}{1+xy}

Complete example

Find the slope of an ellipse at a point

The equation has two local branches near most points, but implicit differentiation finds the tangent slope without solving for either branch first.

x2+4y2=20at(2,2)x^2+4y^2=20\quad\text{at}\quad(2,2)
  1. 1
    Differentiate the equation

    The y-squared term needs the chain-rule factor dy over dx.

    2x+8ydydx=02x+8y\frac{dy}{dx}=0
  2. 2
    Isolate dy over dx

    Move 2x, then divide by 8y.

    dydx=−x4y\frac{dy}{dx}=-\frac{x}{4y}
  3. 3
    Check the point

    The point belongs to the ellipse because 2 squared plus 4 times 2 squared equals 20.

    22+4(22)=202^2+4(2^2)=20
  4. 4
    Evaluate the slope

    Substitute x = 2 and y = 2 only after differentiating.

    dydx∣(2,2)=−28=−14\left.\frac{dy}{dx}\right|_{(2,2)}=-\frac{2}{8}=-\frac14
dydx∣(2,2)=−14\left.\frac{dy}{dx}\right|_{(2,2)}=-\frac14

Verification: On the upper branch y = one half times the square root of 20 minus x squared, explicit differentiation gives -x divided by 2 times that square root. Since the square root equals 2y, this simplifies to -x/(4y).

Avoidable errors

Common mistakes and how to fix them

Missing dy over dx

Problem: Writing the derivative of y squared as 2y.

Why it matters: Along the curve, y changes when x changes, so the chain rule contributes dy over dx.

Better approach: Differentiate y squared as 2y times dy over dx.

Dropping half of the product rule

Problem: Differentiating xy as only y or only x times dy over dx.

Why it matters: Both x and y depend on x in the product-rule calculation.

Better approach: Write d(xy)/dx as y + x times dy over dx.

Substituting a point before differentiating

Problem: Replacing x and y with coordinates in the original equation before finding the slope formula.

Why it matters: The equation then becomes a numerical identity with no local change left to differentiate.

Better approach: Find dy over dx symbolically, confirm the point lies on the curve, and substitute last.

Ignoring a zero denominator

Problem: Reporting a finite slope from a formula whose denominator is zero.

Why it matters: The implicit function condition can fail, and the curve may have a vertical tangent or singular point.

Better approach: Inspect the differentiated equation directly and analyze the point instead of dividing by zero.

Trust the result for the right reasons

Checks, assumptions, and limits

How results are checked

  • Substitute the proposed dy over dx back into the differentiated relation and simplify to an identity.
  • At a regular point, solve for a local explicit branch when practical and compare its derivative.
  • Confirm any evaluation point satisfies the original equation before using it in the slope formula.

When to stop and revise the input

  • The finite slope formula is unsupported where the coefficient of dy over dx is zero unless the point is analyzed separately.
  • Relations with multiple dependent variables, nonsmooth branches, or ambiguous function roles need a more specific setup.
  • Real logarithms, radicals, and denominators retain their original domain restrictions after differentiation.

Common questions

Differentiate an implicit equation FAQ

Why does dy/dx appear when differentiating y?

The equation makes y a function of x along the curve. The chain rule therefore gives d(y)/dx = dy/dx and d(y squared)/dx = 2y times dy/dx.

Do I need to solve the equation for y first?

No. The main advantage of implicit differentiation is that it works directly with the relation, even when isolating y would create several branches or difficult algebra.

What does a zero denominator in dy/dx mean?

It means the displayed finite-slope formula cannot be used there. The point may have a vertical tangent, a singularity, or no regular curve branch, so inspect the original and differentiated equations together.

How do I find the tangent slope at a point?

Differentiate the equation, solve for dy/dx with its restrictions, confirm the point lies on the original curve, and then substitute its coordinates.

Sources and curriculum alignment

This page follows standard introductory mathematics notation and learning sequences. Use these references to continue with a complete course treatment.

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Created by Mathos AI. Methods, conditions, and checks are shown so you can review the mathematical reasoning.