# Implicit Differentiation **A method of differentiating implicitly defined functions.** An [[implicit function]] may be *too complex* to rearrange into an explicit function of one variable; it may be impossible. Thus, to determine the [[derivative]] of the function, it must be *implicitly differentiated*. Implicit differentiation is an application of the [[chain rule]]. ## Implicit functions with two variables In implicit differentiation, every term is differentiated *with respect to one variable*. Terms in which the *variable of differentiation* does not appear will have to be differentiated using the *chain rule*. >[!Explanation]- > As an implicit function *must satisfy* the equation that defines it, the values of its variables are *dependant* on each other; a change in one variable must cause a change in the other. > > If the variables are $x$ and $y$, with $x$ being the *variable of differentiation*, then because $y$ is dependant on $x$ it can be considered a *function of $x$*, i.e. $y=g(x)$, for some function $g$.[^1] > > If $y$ appears in the equation in a *term*, that term can be considered a *function of $y$*, i.e. $f(y)$. As $y$ is already a function of $x$, all terms containing $y$ are *composite functions* of the general form $f(g(x))$. > > As a result, differentiating terms containing $y$ with respect to $x$ must involve the *chain rule*. > > For example, if $y$ is defined as an *implicit function of $x$* by $x^{2}+y^{2}-1=0$, then it is given that $y=g(x)$. The $y^{2}$ term can be considered a function of $y$, i.e. $f(y)=y^{2}$. Thus, $y^{2}$ is actually a *composite function* $f(g(x))=g(x)^{2}$ and the chain rule must be applied to differentiate it. ### Examples >[!Simple implicit function with an expressible explicit form]- > Suppose the derivative $\displaystyle\frac{dy}{dx}$ is to be determined for > $y+x+5=0$ > > Using implicit differentiation begins with differentiating every term with respect to $x$.[^2] > $\begin{align*} > \frac{d}{dx}\;y+\frac{d}{dx}\;x+\frac{d}{dx}\;5&=\frac{d}{dx}\;0 \\ > \frac{dy}{dx}\frac{d}{dy}y+1+0&=0 \\ > \therefore \frac{dy}{dx}&=-1 > \end{align*}$ > which is the same result as obtained by explicit differentiation. >[!Implicit function without an expressible explicit form]- > Suppose the derivative $\displaystyle\frac{dy}{dx}$ is to be determined for > $y^{5}-y=x$ > Using implicit differentiation,[^3] > $\begin{align*} > \frac{d}{dx}\;y^{5}-\frac{d}{dx}\;y&=\frac{d}{dx}\;x \\ > \frac{dy}{dx}\frac{d}{dy}\;y^{5}-\frac{dy}{dx}\frac{d}{dy}\;y&=1 \\ > 5y^{4}\frac{dy}{dx}-\frac{dy}{dx}&=1 \\ > \therefore \frac{dy}{dx}&=\frac{1}{5y^{4}-1} > \end{align*}$ > Clearly, the derivative is *not defined* for $\displaystyle y=\pm\frac{1}{\sqrt[4]{5}}$[^4] which corresponds to the points on the graph where the *tangent is vertical*. > > ![[ImplicitDifferentiationExampleGraph.svg|500]] ## Implicit functions with more than two variables Implicit differentiation of functions containing more than two variables uses the *exact same process* as above, except that every term is [[Partial Derivative|partially differentiated]] instead. >[!Implicit function with three variables]- > Suppose the partial derivative $\displaystyle\frac{\partial z}{\partial x}$ is to be determined for > $x^{3}z^{2}-5xy^{5}z=x^{2}+y^{2}$ > Using implicit differentiation, > $\begin{align*} > 3x^{2}z^{2}+2x^{3}z\frac{\partial z}{\partial x}-5y^{5}z-5xy^{5}\frac{\partial z}{\partial x}&=2x \\ > (2x^{3}z-5xy^{5})\frac{\partial z}{\partial x}&=2x-3x^{2}z^{2}+5y^{5}z \\ > \therefore \frac{\partial z}{\partial x}&=\frac{2x-3x^{2}z^{2}+5y^{5}z}{2x^{3}z-5xy^{5}} > \end{align*}$ [^1]: $g(x)$ *does not* have to be *specified*, as this requires it to be defined *explicitly* which defeats the purpose. [^2]: Using the principles of the explanation, the term $y$ is considered as $f(y)=y$. As $y$ is defined as some function of $x$, that term is a *composite function* and must be differentiated using the chain rule. [^3]: Using the principles of the explanation, the terms $y^{5}$ and $y$ are considered as $f_{1}(y)=y^{5}$ and $f_{2}(y)=y$. The subscript is only used to denote the two different functions of $y$. As $y$ is defined as some function of $x$, the terms $y^{5}$ and $y$ are both *composite functions*. [^4]: The derivative is also not defined for $\displaystyle y=\pm\frac{i}{\sqrt[4]{5}}$.