{{Short description|Mathematical operation in calculus}} {{Calculus}} 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 :<math>F(x,y)=0</math> defines <math>y</math> as a function of <math>x</math>, at least locally, implicit differentiation treats <math>y</math> as a function <math>y(x)</math> and differentiates both sides of the equation with respect to <math>x</math>. The method is an application of the chain rule.<ref name="OpenStax">{{cite book |last1=Strang |first1=Gilbert |last2=Herman |first2=Edwin |title=Calculus Volume 1 |publisher=OpenStax |year=2016 |chapter=3.8 Implicit Differentiation |url=https://openstax.org/books/calculus-volume-1/pages/3-8-implicit-differentiation |isbn=978-1-938168-02-4}}</ref>
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 <math>x^2+y^2=1</math> cannot be represented globally as the graph of a single function <math>y=f(x)</math>, but its upper and lower arcs can each be differentiated implicitly.
The method allows for the computation of the tangent line approximation to <math>y(x)</math>, given only the function <math>F</math> and the point <math>(x_0,y_0)</math> at which the approximation is made, so that no other knowledge of the functional dependence of <math>y</math> on <math>x</math> 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 <math>F</math> be a differentiable function of two variables, and suppose that an equation
:<math>F(x,y)=f(x_0,y_0)</math>
defines <math>y</math> locally as a differentiable function of <math>x</math>, near the point <math>(x,y)=(x_0,y_0)</math>. Applying the differential to both sides gives <math display="block">dF = 0</math> on the curve of interest, and writing the left-hand side in terms of the partial derivatives of <math>F</math>: <math display="block">F_x(x,y)dx + F_y(x,y)dy = 0.</math> Therefore, evaluating at the point <math>(x_0,y_0)</math> gives<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 display="block">\frac{dy}{dx} = - \frac{F_x(x_0,y_0)}{F_y(x_0,y_0)}</math> provided <math>F_y(x_0,y_0)\not=0</math>.
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 <math>F(x,y)=F(x_0,y_0)</math> through the given point <math>(x_0,y_0)</math>. The formula is local, in the sense that it gives only the derivative of the branch of the implicit function <math>y(x)</math> whose graph contains the point <math>(x_0,y_0)</math>.
==Relation with the implicit function theorem== Intuitively, in order to be valid at a point <math>(x_0,y_0)</math> implicit differentiation in the basic formulation requires that the function <math>F</math> must genuinely depend on <math>y</math> near the point <math>(x_0,y_0)</math>, for otherwise the equation <math>F(x,y)=F(x_0,y_0)</math> cannot be solved for <math>y</math> as a function of <math>x</math> at all.
More is actually required: it is clear from the formula that the partial derivative <math>F_y(x_0,y_0)</math> must be non-zero, because it appears in the denominator of the right-hand side of the formula for <math>dy/dx_{(x_0,y_0)}</math>. 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'' <math>y(x)</math> whose graph is contained on the curve <math>F(x,y)=0</math> through the point.
The implicit function theorem supplies the missing justification. It asserts as follows: suppose that <math>F(x,y)</math> is a function such that both partial derivatives exist in a disc around <math>(x,y)=(x_0,y_0)</math>, and are continuous throughout the disc. Then if <math>F_y(x_0,y_0) \ne 0</math>, there is a differentiable function <math>y(x)</math> defined in an open interval containing <math>x=x_0</math>, such that <math>F(x,y(x))=0</math> throughout the interval, and whose derivative is given by the implicit differentiation formula at <math>(x_0,y_0)</math>. The function is, moreover, unique on a sufficiently small interval around <math>x_0</math>.<ref>{{cite book |last=Apostol |first=Tom M. |author-link=Tom M. Apostol |title=Mathematical Analysis |edition=2nd |publisher=Addison-Wesley |year=1974 |isbn=978-0-201-00288-1 |chapter=13. Implicit functions and extremum problems }}</ref>
==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 express {{mvar|y}} explicitly in radicals as a function of {{mvar|x}}, and therefore one cannot find {{math|{{sfrac|''dy''|''dx''}}}} by explicit differentiation of elementary radical expressions. 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>
==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 <math>y(x)</math> at the point of its graph <math>(x_0,y_0)</math>, provided the function <math>F</math> is sufficiently smooth at the point.<ref>{{cite book |last=Goursat |first=Édouard |author-link=Édouard Goursat |translator-last=Hedrick |translator-first=Earle Raymond |title=A Course in Mathematical Analysis |volume=1 |chapter=II. Implicit Functions; Functional Determinants; Change of Variable |publisher=Ginn & Company |location=Boston |year=1904 |pages=35–88 }}</ref>
For example, suppose that the first derivative of <math>y(x)</math> has been found, by solving the equation <math display="block">dF = F_x(x_0,y_0)\,dx + F_y(x_0,y_0)\,dy = 0</math> for <math>dy/dx</math> at <math>(x_0,y_0)</math>.
Applying the second differential to <math>F(x,y) = F(x_0,y_0)</math> then gives <math display="block">d^2 F = 0</math> and expanding the left-hand side gives <math display="block">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^2x + F_y\,d^2y=0.</math> where we have written the second partial derivatives of <math>F</math> with repeated subscripts. Since it is the <math>x</math> derivatives we are after, <math>d^2x=0</math>. Also, substituting the ''known'' expressions for <math>dy</math> at the point <math>(x_0,y_0)</math>, into the formula (that is <math>dy = \frac{dy}{dx}dx</math> where the derivative on the right-hand side is known at the point <math>(x_0,y_0)</math>: <math>dy/dx_{(x_0,y_0)} = -F_x(x_0,y_0)/F_y(x_0,y_0)</math>), we can then arrange terms and solve for the second derivative <math>d^2y/dx^2</math>. That is, from <math display="block">F_{xx} dx^2 + (F_{xy}+F_{yx})\frac{dy}{dx}dx^2 + F_{yy}dy^2 + F_y\,d^2y= 0 </math> we obtain <math display="block"> \frac{d^2y}{dx^2} = - \frac { F_{xx} + (F_{xy}+F_{yx})\frac{dy}{dx}+ F_{yy}\left(\frac{dy}{dx}\right)^2 } { F_{y} } .</math> This is understood as at the point <math>(x_0,y_0)</math>, and requires only that the function <math>F</math> be twice-differentiable at the point, in addition to the hypothesis <math>F_y(x_0,y_0)\ne 0</math>. Second-differentiability at the point is true if the first and second partials of <math>F</math> exist and are continuous in a disc around the point.
In that case, Clairaut's theorem implies that the mixed partials are equal <math>F_{xy}(x_0,y_0)=F_{yx}(x_0,y_0)</math>, and these terms can then be combined to give <math display="block"> \frac{d^2y}{dx^2} = - \frac { F_{xx} - 2F_{xy}\frac{F_x}{F_y} + F_{yy}\left(\frac{F_x}{F_y}\right)^2 } { F_{y} } .</math>
Higher order derviatives (under the requisite smoothness hypotheses) are handled similarly.
==References== {{reflist}}
{{Portal|Mathematics}}
{{Calculus topics}}
Category:Differential calculus