Implicit Differentiation

For an independent variable \(x,\) suppose \(y\) is a dependent variable that depends on \(x;\) i.e. as \(x\) changes, so must \(y\). If the dependence of \(y\) on \(x\) can be described explicitly then we say that \(y\) is a function of \(x\) and we can write \(y = f(x).\) In this case we can also describe how \(y\) changes with respect to \(x\) explicitly by taking the derivative of \(y\) with respect to \(x.\) It’s common to use Newton’s “dot” notation to denote the derivative of dependent variables: \(\dot{y} = f'(x)\) and \(\ddot{y} = f''(x)\) and so on.

However, the relationship between \(x\) and \(y\) cannot always be made explicit; sometimes \(y\) cannot be written as a function of \(x.\) In this case, even if the relationship between \(x\) and \(y\) is only implicit, we can still take an implicit derivative of \(y\) with respect to \(x\) to get an implicit description \(\dot{y}\) for how \(y\) is changing with respect to \(x\). The derivative \(\dot{y}\) can always be described explicitly in terms of \(x\) and \(y.\)

Example

Suppose \(x\) is an independent variable and \(y\) is a dependent variable that depends on \(x\) such that \(\sin(y) = xy\). Then

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\bigl( \sin(y) \bigr) = \frac{\mathrm{d}}{\mathrm{d}x}\bigl( xy \bigr) \quad\implies\quad {\color{SteelBlue}\underbrace{{\color{Black}\cos(y)\dot{y}}}_{\text{chain rule}} {\color{black}\;\;=} \underbrace{{\color{Black}x\dot{y} + y}}_{\text{product rule}}} \quad\implies\quad \dot{y} = \frac{y}{\cos(y)-x} \)

Note that if the independent variable represents time, it’s common to use the variable \(t\) instead of \(x.\)