*In this section, we will learn that even when $y$ is not explicitly expressed in terms of $x$, we can apply differentiation rules to find $dy/dx$. This is called implicit differentiation. *

In most of the functions that we have encountered, the dependent variable, \(y\), has been explicitly expressed in terms of the independent variable, \(x\), as in \[y=\frac{\sin x}{x},\qquad\text{or}{\qquad y=x\sqrt{x^{2}+1}.}\] 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 \[x^{2}+y^{2}=4,\] or \[x^{3}-3xy+y^{3}=0.\] 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 Figure 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.

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, Figure 3).

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 \(dy/dx\).

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

## Higher Order Derivatives Using Implicit Differentiation

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