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In calculus, implicit differentiation is a method for finding the derivative of a function that is defined by an equation rather than by an explicit formula. If an equation such as

F(x,y)=0

defines $y$ as a function of $x$ , at least locally, implicit differentiation treats $y$ as a function $y(x)$ and differentiates both sides of the equation with respect to $x$ . The method is an application of the chain rule.^[1]

Implicit differentiation is useful when solving explicitly for one variable is inconvenient, produces several branches, or is not possible in elementary terms. For example, the circle $x^{2}+y^{2}=1$ cannot be represented globally as the graph of a single function $y=f(x)$ , but its upper and lower arcs can each be differentiated implicitly.

The method allows for the computation of the tangent line approximation to $y(x)$ , given only the function $F$ and the point $(x_{0},y_{0})$ at which the approximation is made, so that no other knowledge of the functional dependence of $y$ on $x$ is needed. Tangent approximations to surfaces and higher-dimensional manifolds given by an equation or system of equations can be found by similar methods.

Basic formula

Let $F$ be a differentiable function of two variables, and suppose that an equation

F(x,y)=f(x_{0},y_{0})

defines $y$ locally as a differentiable function of $x$ , near the point $(x,y)=(x_{0},y_{0})$ . Applying the differential to both sides gives $dF=0$ on the curve of interest, and writing the left-hand side in terms of the partial derivatives of $F$ : $F_{x}(x,y)dx+F_{y}(x,y)dy=0.$ Therefore, evaluating at the point $(x_{0},y_{0})$ gives^[2]^: §11.5 ${\frac {dy}{dx}}=-{\frac {F_{x}(x_{0},y_{0})}{F_{y}(x_{0},y_{0})}}$ provided $F_{y}(x_{0},y_{0})\not =0$ .

The derivative on the left-hand side of the formula can be interpreted as the slope of the tangent line to the curve defined implicitly by the equation $F(x,y)=F(x_{0},y_{0})$ through the given point $(x_{0},y_{0})$ . The formula is local, in the sense that it gives only the derivative of the branch of the implicit function $y(x)$ whose graph contains the point $(x_{0},y_{0})$ .

Relation with the implicit function theorem

Intuitively, in order to be valid at a point $(x_{0},y_{0})$ implicit differentiation in the basic formulation requires that the function $F$ must genuinely depend on $y$ near the point $(x_{0},y_{0})$ , for otherwise the equation $F(x,y)=F(x_{0},y_{0})$ cannot be solved for $y$ as a function of $x$ at all.

More is actually required: it is clear from the formula that the partial derivative $F_{y}(x_{0},y_{0})$ must be non-zero, because it appears in the denominator of the right-hand side of the formula for $dy/dx_{(x_{0},y_{0})}$ . While this fact allows one to compute a numerical value, it is not obvious that the numerical value has the advertised meaning: namely, that it be the derivative of a function $y(x)$ whose graph is contained on the curve $F(x,y)=0$ through the point.

The implicit function theorem supplies the missing justification. It asserts as follows: suppose that $F(x,y)$ is a function such that both partial derivatives exist in a disc around $(x,y)=(x_{0},y_{0})$ , and are continuous throughout the disc. Then if $F_{y}(x_{0},y_{0})\neq 0$ , there is a differentiable function $y(x)$ defined in an open interval containing $x=x_{0}$ , such that $F(x,y(x))=0$ throughout the interval, and whose derivative is given by the implicit differentiation formula at $(x_{0},y_{0})$ . The function is, moreover, unique on a sufficiently small interval around $x_{0}$ .^[3]

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

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 $y \geq 0$ 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 express $y$ explicitly in radicals as a function of $x$ , and therefore one cannot find $⁠ dy / dx ⁠$ by explicit differentiation of elementary radical expressions. 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

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

Higher order derivatives

The method of implicit differentiation can be applied to find higher-order derivatives such as the second derivative of the implicitly-defined function $y(x)$ at the point of its graph $(x_{0},y_{0})$ , provided the function $F$ is sufficiently smooth at the point.^[4]

For example, suppose that the first derivative of $y(x)$ has been found, by solving the equation $dF=F_{x}(x_{0},y_{0})\,dx+F_{y}(x_{0},y_{0})\,dy=0$ for $dy/dx$ at $(x_{0},y_{0})$ .

Applying the second differential to $F(x,y)=F(x_{0},y_{0})$ then gives $d^{2}F=0$ and expanding the left-hand side gives $d(F_{x}\,dx+F_{y}\,dy)=F_{xx}dx^{2}+F_{yx}dx\,dy+F_{xy}dy\,dx+F_{yy}dy^{2}+F_{x}\,d^{2}x+F_{y}\,d^{2}y=0.$ where we have written the second partial derivatives of $F$ with repeated subscripts. Since it is the $x$ derivatives we are after, $d^{2}x=0$ . Also, substituting the known expressions for $dy$ at the point $(x_{0},y_{0})$ , into the formula (that is $dy={\frac {dy}{dx}}dx$ where the derivative on the right-hand side is known at the point $(x_{0},y_{0})$ : $dy/dx_{(x_{0},y_{0})}=-F_{x}(x_{0},y_{0})/F_{y}(x_{0},y_{0})$ ), we can then arrange terms and solve for the second derivative $d^{2}y/dx^{2}$ . That is, from $F_{xx}dx^{2}+(F_{xy}+F_{yx}){\frac {dy}{dx}}dx^{2}+F_{yy}dy^{2}+F_{y}\,d^{2}y=0$ we obtain ${\frac {d^{2}y}{dx^{2}}}=-{\frac {F_{xx}+(F_{xy}+F_{yx}){\frac {dy}{dx}}+F_{yy}\left({\frac {dy}{dx}}\right)^{2}}{F_{y}}}.$ This is understood as at the point $(x_{0},y_{0})$ , and requires only that the function $F$ be twice-differentiable at the point, in addition to the hypothesis $F_{y}(x_{0},y_{0})\neq 0$ . Second-differentiability at the point is true if the first and second partials of $F$ exist and are continuous in a disc around the point.

In that case, Clairaut’s theorem implies that the mixed partials are equal $F_{xy}(x_{0},y_{0})=F_{yx}(x_{0},y_{0})$ , and these terms can then be combined to give ${\frac {d^{2}y}{dx^{2}}}=-{\frac {F_{xx}-2F_{xy}{\frac {F_{x}}{F_{y}}}+F_{yy}\left({\frac {F_{x}}{F_{y}}}\right)^{2}}{F_{y}}}.$

Higher order derviatives (under the requisite smoothness hypotheses) are handled similarly.

References

^ Strang, Gilbert; Herman, Edwin (2016). “3.8 Implicit Differentiation”. Calculus Volume 1. OpenStax. ISBN 978-1-938168-02-4.
^ Stewart, James (1998). Calculus Concepts And Contexts. Brooks/Cole Publishing Company. ISBN 0-534-34330-9.
^ Apostol, Tom M. (1974). “13. Implicit functions and extremum problems”. Mathematical Analysis (2nd ed.). Addison-Wesley. ISBN 978-0-201-00288-1.
^ Goursat, Édouard (1904). “II. Implicit Functions; Functional Determinants; Change of Variable”. A Course in Mathematical Analysis. Vol. 1. Translated by Hedrick, Earle Raymond. Boston: Ginn & Company. pp. 35–88.

[rdp-we-cite_note-OpenStax-1] Strang, Gilbert; Herman, Edwin (2016). “3.8 Implicit Differentiation”. Calculus Volume 1. OpenStax. ISBN 978-1-938168-02-4.

[rdp-we-cite_note-Stewart1998-2] Stewart, James (1998). Calculus Concepts And Contexts. Brooks/Cole Publishing Company. ISBN 0-534-34330-9.

[rdp-we-cite_note-3] Apostol, Tom M. (1974). “13. Implicit functions and extremum problems”. Mathematical Analysis (2nd ed.). Addison-Wesley. ISBN 978-0-201-00288-1.

[rdp-we-cite_note-4] Goursat, Édouard (1904). “II. Implicit Functions; Functional Determinants; Change of Variable”. A Course in Mathematical Analysis. Vol. 1. Translated by Hedrick, Earle Raymond. Boston: Ginn & Company. pp. 35–88.

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