Implicit differentiation

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 R(x, y) = 0, it is not generally possible to solve it explicitly for y and then differentiate it. Instead, one can totally differentiate R(x, y) = 0 with respect to x and y and then solve the resulting linear equation for ⁠dy/dx⁠, 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 R(x, y) = 0, 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 R_x and R_y 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 R(x, y) = 0:

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

hence

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

which, when solved for ⁠dy/dx⁠, 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 ⁠dy/dx⁠ = −1.

Alternatively, one can totally differentiate the original equation:

$\begin{array}{cc}\frac{dy}{dx}+\frac{dx}{dx}+\frac{d}{dx}(5)& =0;\\ [6px]\frac{dy}{dx}+1+0& =0.\end{array}$

Solving for ⁠dy/dx⁠ 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 y < 0.

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 ⁠dy/dx⁠ by explicit differentiation. Using the implicit method, ⁠dy/dx⁠ can be obtained by differentiating the equation to obtain

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

where ⁠dx/dx⁠ = 1. Factoring out ⁠dy/dx⁠ shows that

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

which yields the result

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

which is defined for

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