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Incalculus,implicit differentiation is a method of finding thederivative of animplicit function using thechain rule.To differentiate an implicit functiony(x), defined by an equationR(x,y) = 0, it isnot generally possible tosolve it explicitly fory and then differentiate it. Instead, one cantotally differentiateR(x,y) = 0 with respect tox andy and then solve the resulting linear equation for⁠dy/dx⁠, to get the derivative explicitly in terms ofx andy. 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.

IfR(x,y) = 0, the derivative of the implicit functiony(x) is given by^{[1]: §11.5}

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

whereR_x andR_y indicate thepartial derivatives ofR with respect tox andy.

The above formula comes from using thegeneralized chain rule to obtain thetotal derivative — with respect tox — of both sides ofR(x,y) = 0:

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

hence

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

which, when solved for⁠dy/dx⁠, gives the expression above.

Consider

${\displaystyle y+x+5=0.}$

This equation is easy to solve fory, giving

${\displaystyle y=-x-5,}$

where the right side is the explicit form of the functiony(x). Differentiation then gives⁠dy/dx⁠ = −1.

Alternatively, one can totally differentiate the original equation:

Solving for⁠dy/dx⁠ gives

${\displaystyle \frac{dy}{dx}=-1,}$

the same answer as obtained previously.

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

${\displaystyle x^4+2y^2=8.}$

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

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

and then differentiate this function. This creates two derivatives: one fory ≥ 0 and another fory < 0.

It is substantially easier to implicitly differentiate the original equation:

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

giving

${\displaystyle \frac{dy}{dx}=\frac{-4x^3}{4y}=-\frac{x^3}{y}.}$

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

${\displaystyle y^5-y=x.}$

It is impossible toalgebraically expressy explicitly as a function ofx, and therefore one cannot find⁠dy/dx⁠ by explicit differentiation. Using the implicit method,⁠dy/dx⁠ can be obtained by differentiating the equation to obtain

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

where⁠dx/dx⁠ = 1. Factoring out⁠dy/dx⁠ shows that

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

which yields the result

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

which is defined for

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

• Differential calculus

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