CCNY, Fall 2017, Math 19500 Precalculus
CCNY, Spring 2018, Math 20200 Calculus II

Implicit differentiation in multiple variables

In single variable calculus, one is taught the method to compute the tangent line of a curves defined by equation, namely implicit differentiation.

For example, consider the unit circle \(C : x^2 + y^2 = 1\). Applying \(d\) to both sides and use chain rule formally, one derives \(2x \, dx + 2y \, dy = 0\) whence it follows that \(\frac{dy}{dx} = -\frac{x}{y}\). What this computation really means is that given an explicit point, say \(P = (\frac35, \frac45)\) on the circle, the slope of the tangent line to the circle at that point is \(-\frac{x_P}{y_P} = -\frac{3/5}{4/5} = -\frac34\). From here, it is easy to write down the equation of the tangent line: \[y = -\frac34 (x - \frac35) + \frac45.\]

So how does one generalize this method to higher dimension, for instance, a surface defined by an equation such as \[S : xz + y e^z = z^2\] in 3D space or a curve defined by two equations?

First of all, observe that the tangent to a surface in 3D is not a line but a plane whereas the tangent to a curve is still a line. In general, the tangent space should have the same dimension as the object it is tangent to.

Secondly, observe that a plane in 3D can be identified by a point and a normal vector. Alternatively, we can use a point and two lines lying on the plane.

Finally, what does it mean to be tangent? The tangent plane is supposed to be the "best plane" that looks like the given surface at the point. This is the actual meaning of derivatives: to provide linear approximation to functions.

\( \newcommand{ppf}[2]{\frac{\partial #1}{\partial #2}} \newcommand{pp}[1]{\frac{\partial}{\partial #1}} \) In the example above, apply differential operators \(\pp{x}\) and \(\pp{y}\) to both sides of the equation, we get \[ \begin{align*} \pp{x} &:& z + x \ppf{z}{x} + y e^z \ppf{z}{x} &= 2z \ppf{z}{x}\\ \pp{y} &:& x \ppf{z}{y} + e^z + y e^z \ppf{z}{x} &= 2z \ppf{z}{y}\\ \end{align*} \] whence \[ \begin{align*} \ppf{z}{x} &= \frac{z}{2z - x - ye^z}\\ \ppf{z}{y} &= \frac{e^z}{2z - x - ye^z}\\ \end{align*} \]

Solving inequalities