Implicit differentiation

---

title: "Implicit differentiation" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["differential-calculus"] description: "Mathematical operation in calculus" topic_path: "science/mathematics" source: "https://en.wikipedia.org/wiki/Implicit_differentiation" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::summary Mathematical operation in calculus ::

In calculus, implicit differentiation is a method of finding the derivative of an implicit function using the chain rule. To differentiate an implicit function y(x), defined by an equation , it is not generally possible to solve it explicitly for y and then differentiate it. Instead, one can totally differentiate with respect to x and y and then solve the resulting linear equation for , to get the derivative explicitly in terms of x and y. Even when it is possible to explicitly solve the original equation, the formula resulting from total differentiation is, in general, much simpler and easier to use.

Formulation

If , the derivative of the implicit function y(x) is given by

:\frac{dy}{dx} = -\frac{,\frac{\partial R}{\partial x},}{\frac{\partial R}{\partial y}} = -\frac {R_x}{R_y} ,,

where Rx and Ry indicate the partial derivatives of R with respect to x and y.

The above formula comes from using the generalized chain rule to obtain the total derivative — with respect to x — of both sides of :

:\frac{\partial R}{\partial x} \frac{dx}{dx} + \frac{\partial R}{\partial y} \frac{dy}{dx} = 0 ,,

hence

:\frac{\partial R}{\partial x} + \frac{\partial R}{\partial y} \frac{dy}{dx} =0 ,,

which, when solved for , gives the expression above.

Examples

Example 1

Consider

:y + x + 5 = 0 ,.

This equation is easy to solve for y, giving

:y = -x - 5 ,,

where the right side is the explicit form of the function y(x). Differentiation then gives .

Alternatively, one can totally differentiate the original equation:

:\begin{align} \frac{dy}{dx} + \frac{dx}{dx} + \frac{d}{dx}(5) &= 0 , ; \[6px] \frac{dy}{dx} + 1 + 0 &= 0 ,. \end{align}

Solving for gives

:\frac{dy}{dx} = -1 ,,

the same answer as obtained previously.

Example 2

An example of an implicit function for which implicit differentiation is easier than using explicit differentiation is the function y(x) defined by the equation

: x^4 + 2y^2 = 8 ,.

To differentiate this explicitly with respect to x, one has first to get

:y(x) = \pm\sqrt{\frac{8 - x^4}{2}} ,,

and then differentiate this function. This creates two derivatives: one for y ≥ 0 and another for {{math|y

It is substantially easier to implicitly differentiate the original equation:

:4x^3 + 4y\frac{dy}{dx} = 0 ,,

giving

:\frac{dy}{dx} = \frac{-4x^3}{4y} = -\frac{x^3}{y} ,.

Example 3

Often, it is difficult or impossible to solve explicitly for y, and implicit differentiation is the only feasible method of differentiation. An example is the equation

:y^5-y=x ,.

It is impossible to algebraically express y explicitly as a function of x, and therefore one cannot find by explicit differentiation. Using the implicit method, can be obtained by differentiating the equation to obtain

:5y^4\frac{dy}{dx} - \frac{dy}{dx} = \frac{dx}{dx} ,,

where . Factoring out shows that

:\left(5y^4 - 1\right)\frac{dy}{dx} = 1 ,,

which yields the result

:\frac{dy}{dx}=\frac{1}{5y^4-1} ,,

which is defined for

:y \ne \pm\frac{1}{\sqrt[4]{5}} \quad \text{and} \quad y \ne \pm \frac{i}{\sqrt[4]{5}} ,.

References

References

1. Stewart, James. (1998). "Calculus Concepts And Contexts". Brooks/Cole Publishing Company.

::callout[type=info title="Wikipedia Source"] This article was imported from Wikipedia and is available under the Creative Commons Attribution-ShareAlike 4.0 License. Content has been adapted to SurfDoc format. Original contributors can be found on the article history page. ::