Editing Implicit function (section)

==Implicit differentiation==
In [[calculus]], a method called '''implicit differentiation''' makes use of the [[chain rule]] to differentiate implicitly defined functions.

To differentiate an implicit function {{math|''y''(''x'')}}, defined by an equation {{math|1=''R''(''x'', ''y'') = 0}}, it is not generally possible to solve it explicitly for {{mvar|y}} and then differentiate. Instead, one can [[total differentiation|totally differentiate]] {{math|1=''R''(''x'', ''y'') = 0}} with respect to {{mvar|x}} and {{mvar|y}} and then solve the resulting linear equation for {{math|{{sfrac|''dy''|''dx''}}}} to explicitly get the derivative in terms of {{mvar|x}} and {{mvar|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.

===Examples===
==== Example 1 ====
Consider

:<math>y + x + 5 = 0 \,.</math>

This equation is easy to solve for {{mvar|y}}, giving

:<math>y = -x - 5 \,,</math>

where the right side is the explicit form of the function {{math|''y''(''x'')}}. Differentiation then gives {{math|1={{sfrac|''dy''|''dx''}} = −1}}.

Alternatively, one can totally differentiate the original equation:

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

Solving for {{math|{{sfrac|''dy''|''dx''}}}} gives

:<math>\frac{dy}{dx} = -1 \,,</math>

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 {{math|''y''(''x'')}} defined by the equation

:<math> x^4 + 2y^2 = 8 \,.</math>

To differentiate this explicitly with respect to {{mvar|x}}, one has first to get

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

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

It is substantially easier to implicitly differentiate the original equation:

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

giving

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

==== Example 3 ====
Often, it is difficult or impossible to solve explicitly for {{mvar|y}}, and implicit differentiation is the only feasible method of differentiation. An example is the equation

:<math>y^5-y=x \,.</math>

It is impossible to [[algebraic expression|algebraically express]] {{mvar|y}} explicitly as a function of {{mvar|x}}, and therefore one cannot find {{math|{{sfrac|''dy''|''dx''}}}} by explicit differentiation. Using the implicit method, {{math|{{sfrac|''dy''|''dx''}}}} can be obtained by differentiating the equation to obtain

:<math>5y^4\frac{dy}{dx} - \frac{dy}{dx} = \frac{dx}{dx} \,,</math>

where {{math|1={{sfrac|''dx''|''dx''}} = 1}}. Factoring out {{math|{{sfrac|''dy''|''dx''}}}} shows that

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

which yields the result

:<math>\frac{dy}{dx}=\frac{1}{5y^4-1} \,,</math>

which is defined for

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

===General formula for derivative of implicit function===
If {{math|1=''R''(''x'', ''y'') = 0}}, the derivative of the implicit function {{math|''y''(''x'')}} is given by<ref name="Stewart1998">{{cite book | last = Stewart | first = James | title = Calculus Concepts And Contexts | publisher = Brooks/Cole Publishing Company | year = 1998 | isbn = 0-534-34330-9 | url-access = registration | url = https://archive.org/details/calculusconcepts00stew }}</ref>{{rp|§11.5}}

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

where {{math|''R<sub>x</sub>''}} and {{math|''R<sub>y</sub>''}} indicate the [[partial derivative]]s of {{mvar|R}} with respect to {{mvar|x}} and {{mvar|y}}.

The above formula comes from using the [[Chain rule#Multivariable case|generalized chain rule]] to obtain the [[total derivative]] — with respect to {{mvar|x}} — of both sides of {{math|1=''R''(''x'', ''y'') = 0}}:

:<math>\frac{\partial R}{\partial x} \frac{dx}{dx} + \frac{\partial R}{\partial y} \frac{dy}{dx} = 0 \,,</math>

hence

:<math>\frac{\partial R}{\partial x} + \frac{\partial R}{\partial y} \frac{dy}{dx} =0 \,,</math>

which, when solved for {{math|{{sfrac|''dy''|''dx''}}}}, gives the expression above.