In most of the functions that we have encountered, the dependent variable, \(y\), has been explicitly expressed in terms of the independent variable, \(x\). For example, \[y=\frac{\sin x}{x},\qquad\text{or}{\qquad y=x\sqrt{x^{2}+1}.}\] In general, \(y\) is an explicit function of \(x\) if \[y=f(x).\] However, sometimes we have to deal with equations of the form \[F(x,y)=0,\] where the equation is not solved for \(y\) such as

\[x^{2}+y^{2}=4,\] and \[x^{3}-3xy+y^{3}=0.\] In such cases, we say \(y\) is implicitly expressed in terms of \(x\); if a value of \(x\) is given, a value or values of \(y\) may be determined. Sometimes, we can solve these equations for \(y\) in terms of \(x\), thereby \(y\) becomes an explicit function (or perhaps several functions) of \(x\). For example, we can solve the equation \(x^{2}+y^{2}=4\) (this is the equation of a circle of radius 2 centered at the origin) for \(y\) and obtain: \[y=\sqrt{4-x^{2}},\qquad\text{and}{\qquad y=-\sqrt{4-x^{2}}}.\] We may state that the equation \(x^{2}+y^{2}=4\) implicitly defines two functions
\[f(x)=\sqrt{4-x^{2}},\qquad\text{and}{\qquad g(x)=-\sqrt{4-x^{2}}}.\]
The graph of \(f\) is the upper semicircle and the graph of \(g\) is the lower semicircle (see Fig.1).

(a) Graph of \(x^{2}+y^{2}=4\) (b) Graph of \(y=\sqrt{4-x^{2}}\) (c) Graph of \(y=-\sqrt{4-x^{2}}\)
Figure 1

If we can solve explicitly for \(y\), then we can differentiate it as before.

Example 1

If\(x^{2}+y^{2}=4\), find \(\dfrac{dy}{dx}\).


We discussed that \(y\) can be written as \(y=f(x)=\sqrt{4-x^{2}}\) or \(y=g(x)=-\sqrt{4-x^{2}}\). We can easily compute the derivatives of \(f\) and \(g\) by the chain rule:
\[\begin{align} f'(x) & =\frac{d}{dx}\sqrt{4-x^{2}}\\ & =\frac{d}{du}(\sqrt{u})\frac{du}{dx}\tag{${u=4-x^{2}}$}\\ & =\frac{1}{2\sqrt{u}}(-2x)\\ & =\frac{-2x}{2\sqrt{4-x^{2}}}\\ & =-\frac{x}{f(x)}\\ & =-\frac{x}{y},\end{align}\]
whenever \(y=f(x)\neq0\). Similarly we can show \[g'(x)=-\frac{-x}{\sqrt{4-x^{2}}}=-\frac{x}{g(x)}=-\frac{x}{y},\] whenever \(y=g(x)\neq0\). So in general, we can write
\[\frac{dy}{dx}=-\frac{x}{y}\qquad\text{if }y\neq0.\]

Sometimes it is very difficult or even impossible to solve an implicit relation for \(y\). For example, solving the equation \(x^{3}-3xy+y^{3}=0\) for \(y\) in terms of \(x\) is difficult (computer algebra systems such as Mathematica, Maple, and Sympy can solve this equation for \(y\), but the expressions that they give are complicated and long). This equation represents a curve that is called the folium of Descartes (see Figure 2), and implicitly defines infinitely many functions (for example see, Fig. 3).

Figure 2. The folium of Descartes with its asymptote (dashed line)


Figure 3. Graphs of three functions implicitly defined by x^{3}+y^{3}-3xy=0.

If we assume that \(y\) can be defined as one or more differentiable functions of \(x\), we can apply the chain rule to find \(dy/dx\) directly without solving the equation. In this method that is known as implicit differentiation, we differentiate both sides of an equation with respect to \(x\) and treat \(y\) as a differentiable function of \(x\). Then, we try to solve for \(dy/dx\). We will illustrate this method through resolving Example 1.

Example 2

If\(x^{2}+y^{2}=4\), find \(dy/dx\) using implicit differentiation.


We assume \(dy/dx\) exists and apply \(\dfrac{d}{dx}\) to both sides of the equation:
\[\begin{align} \frac{d}{dx}(x^{2}+y^{2}) & =\frac{d}{dx}4\\ \frac{d}{dx}(x^{2})+\frac{d}{dx}y^{2} & =0\tag{Recall that the derivative of a constant is zero}\\ 2x+2y\frac{dy}{dx} & =0.\tag{i}\end{align}\]
For the last step, we used the chain rule:
We can solve Equation (i) for \(dy/dx\) and obtain:

  • Note that in implicit differentiation only those values of the variables which satisfy the original relation can be substituted in the derivative.
Example 3

Find the equation of the tangent line to the curve described by \(x^{3}+y^{3}-3xy=0\) at \((\sqrt[3]{2},\sqrt[3]{4})\).


First of all, we note that \(x=\sqrt[3]{2}\) and \(y=\sqrt[3]{4}\) satisfy \(x^{3}+y^{3}-3xy=0\). Differentiating with respect to \(x\), we obtain:
\[\begin{align} \frac{d}{dx}(x^{3})+\frac{d}{dx}\left(y^{3}\right)-3\frac{d}{dx}(xy) & =0,\\ 3x^{2}+3y^{2}\frac{dy}{dx}-3y-3x\frac{dy}{dx} & =0,\\ \frac{dy}{dx}\left(3y^{2}-3x\right) & =3(y-x^{2})\\ \frac{dy}{dx} & =\frac{y-x^{2}}{y^{2}-x}.\end{align}\]
The slope of the tangent at \((\sqrt[3]{2},\sqrt[3]{4})\) is thus
This shows the tangent to this curve is horizontal and its equation becomes \(y=\sqrt[3]{4}\). The following figure shows the curve described by the equation \(x^{3}+y^{3}-3xy=0\) and its tangent at the point \((\sqrt[3]{2},\sqrt[3]{4})\approx(1.26,1.587)\).

Figure 4. The curve described by \(x^{3}+y^{3}-3xy=0\) and its tangent at \((\sqrt[3]{2},\sqrt[3]{4})\).
Example 4

If \(x^{2}-3xy+2y^{2}=3\), find \(\dfrac{dy}{dx}\).


Here we can solve for \(y\) (because it is a quadratic equation in terms of \(y\)), but it is better if we assume that \(y’\) exists and apply \(\dfrac{d}{dx}\) to both sides of the equation:
\[\begin{align} \frac{d}{dx}(x^{2}-3xy+2y^{2}) & =\frac{d}{dx}3\\ \frac{d}{dx}(x^{2})-3\frac{d}{dx}(xy)+2\frac{d}{dx}(y^{2}) & =0\\ 2x-3\left(y\frac{dx}{dx}+x\frac{dy}{dx}\right)+2\times2y\frac{dy}{dx} & =0\\ 2x-3y-3x\frac{dy}{dx}+4y\frac{dy}{dx} & =0\end{align}\]

thus \[\frac{dy}{dx}(4y-3x)=3y-2x,\] and finally

Derivatives of higher order

Higher order derivatives can also be obtained by application of the implicit differentiation. We will illustrate how to do that through the following example.

Example 5

If \(x^{2}+y^{2}=4\), find \(\dfrac{d^{2}y}{dx^{2}}\).


In Examples 1 and 2, we showed \[\frac{dy}{dx}=-\frac{x}{y}.\] We now apply the quotient rule to find \(y^{\prime\prime}\).
\[\begin{align} \frac{d^{2}y}{dx^{2}} & =\frac{d}{dx}\left(-\frac{x}{y}\right)\\ & =-\frac{\frac{dx}{dx}\cdot y-x\cdot\frac{dy}{dx}}{y^{2}}\\ & =-\frac{y-x\cdot\left(-\frac{x}{y}\right)}{y^{2}}\tag{$\text{substitute }{\frac{dy}{dx}=-x/y}$}\\ & =-\frac{y+\frac{x^{2}}{y}}{y^{2}}\\ & =-\frac{y^{2}+x^{2}}{y^{3}}\\ & =-\frac{4}{y^{3}}\tag{$\text{because } {x^{2}+y^{2}=4}$}\end{align}\]

Example 6

If \(y^{3}-xy-1=0\), find \(\dfrac{d^{2}y}{dx^{2}}\).


First we need to find \(dy/dx\). In doing so, we use implicit differentiation.
\[\begin{align} 3y^{2}\frac{dy}{dx}-y-x\frac{dy}{dx}-0 & =0\\ \frac{dy}{dx}\left(3y^{2}-x\right) & =y\end{align}\]

\[\Rightarrow\frac{dy}{dx}=\frac{y}{3y^{2}-x}.\] Then, we apply the quotient rule to find \(y^{\prime\prime}\).
\[\begin{align} \frac{d^{2}y}{dx^{2}} & =\frac{\frac{dy}{dx}\cdot(3y^{2}-x)-y\frac{d(3y^{2}-x)}{dx}}{(3y^{2}-x)^{2}}\\ & =\frac{\dfrac{y}{3y^{2}-x}\cdot(3y^{2}-x)-y\cdot\left(6y\dfrac{dy}{dx}-1\right)}{(3y^{2}-x)^{2}}\\ & =\frac{y-y\left(\dfrac{6y^{2}}{3y^{2}-x}-1\right)}{(3y^{2}-x)^{2}}\\ & =\frac{y(3y^{2}-x)-y\left(6y^{2}-(3y^{2}-x)\right)}{(3y^{2}-x)^{3}}\\ & =\frac{-2xy}{(2y^{2}-x)^{3}}.\end{align}\]

Example 7

Given \(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\) (this is the equation of an ellipse), find \(\dfrac{d^{2}y}{dx^{2}}\).


First we need to find \(dy/dx\). Using implicit differentiation, we obtain:
\[\begin{align} \frac{1}{a^{2}}\frac{dx^{2}}{dx}+\frac{1}{b^{2}}\frac{dy^{2}}{dx} & =\frac{d}{dx}(1),\\ \frac{2}{a^{2}}x+\frac{2}{b^{2}}y\frac{dy}{dx} & =0,\end{align}\]
\[\Rightarrow\frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}.\] Now we can apply the quotient rule to find \(d^{2}y/dx^{2}\):
\[\begin{align} \frac{d^{2}y}{dx^{2}} & =-\frac{b^{2}}{a^{2}}\frac{\dfrac{dx}{dx}\cdot y-x\cdot\dfrac{dy}{dx}}{y^{2}}\\ & =-\frac{b^{2}}{a^{2}}\dfrac{y-x\left(\dfrac{b^{2}x}{a^{2}y}\right)}{y^{2}}\\ & =-\frac{b^{2}}{a^{4}}\frac{a^{2}y^{2}+b^{2}x^{2}}{y^{3}}\end{align}\]
If we multiply both sides of \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) by \(a^{2}b^{2}\), we conclude that \(b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}\). Thus
\[\begin{align} \frac{d^{2}y}{dx^{2}} & =-\frac{b^{2}}{a^{4}}\frac{a^{2}b^{2}}{y^{3}}\\ & =-\frac{b^{4}}{a^{2}y^{3}}.\end{align}\]
which shows \(f\)’ is not continuous at \(x=0\).

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