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 Template:Em generally possible to solve it explicitly for Template:Mvar and then differentiate it. Instead, one can totally differentiate R(x, y) = 0 with respect to Template:Mvar and Template:Mvar and then solve the resulting linear equation for Template:Sfrac, to get the derivative explicitly in terms of Template:Mvar and Template:Mvar. 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 byTemplate:Rp

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

where R_x and R_y indicate the partial derivatives of Template:Mvar with respect to Template:Mvar and Template:Mvar.

The above formula comes from using the generalized chain rule to obtain the total derivative — with respect to Template:Mvar — 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 Template:Sfrac, gives the expression above.

Examples

Example 1

Consider

$y+x+5=0.$

This equation is easy to solve for Template:Mvar, giving

$y=-x-5,$

where the right side is the explicit form of the function y(x). Differentiation then gives Template:Sfrac = −1.

Alternatively, one can totally differentiate the original equation:

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

Solving for Template:Sfrac 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 Template:Mvar, 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 Template:Mvar, 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 Template:Mvar explicitly as a function of Template:Mvar, and therefore one cannot find Template:Sfrac by explicit differentiation. Using the implicit method, Template:Sfrac can be obtained by differentiating the equation to obtain

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

where Template:Sfrac = 1. Factoring out Template:Sfrac 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

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