Implicit Derivatives

Section 1.2 Implicit Derivatives

Subsection 1.2.1 Derivatives of Loci

In Calculus I, I calculated slopes of tangent lines to graphs of functions. Graphs are loci of the form \(y = f(x)\text{,}\) and \(f^\prime(x)\) gave the slope of the tangent line. For arbitrary loci in \(\RR^2\text{,}\) such as the conics, tangent lines can also be defined. I need a refinement of the derivative techniques to find their slopes, since most loci are not graphs of functions. The new technique is implicit differentiation.

Since I know how to differentiate functions, I will pretend that a locus is (at least locally) a graph of a function. In a locus of \(x\) and \(y\text{,}\) I will pretend that \(y\) is a function of \(x\text{.}\) I will the use this pretense to differentiate the expression of the locus.

Differentiating expression in \(x\) is easy: I just differentiate. Differentiating expressions in \(y\) (or with both variables) needs to account for the fact that I am pretending \(y\) is a function of \(x\text{.}\) If I differentiate \(y^2\text{,}\) I have to treat it as a composition: some function \(y(x)\) which is the inside function, then then the function square \((y(x))^2\) as the outside. As a composition, this is always a chain rule calculation. I get \(2y\) as the derivative of the outside, but I need to multiply by \(\frac{dy}{dx}\text{,}\) the derivative of the inside. I don’t know what this is, so I just leave it as \(\frac{dy}{dx}\text{.}\) I’ll eventually solve for this, since \(\frac{dy}{dx}\) is, in fact, precisely what I want: the slope of the tangent line to the locus.

Figure 1.2.1. Tangent Lines to \(x^4 + y^4 =1\)

Example 1.2.2.

Take the locus of \(x^4 + y^4 = 16\text{,}\) as in Figure 1.2.1. I will pretend that \(y\) is locally a function of \(x\text{:}\) for any expressions in \(x\text{,}\) I differentiate normally. For expressions in \(y\text{,}\) I use the chain rule, due to the pretense that \(y\) is a function of \(x\text{.}\) Therefore, \(y^4\) has a derivative of the outside (\(4y^3\)) and a derivative of the inside (\(\frac{dy}{dx}\)). Then I solve for \(\frac{dy}{dx}\text{.}\)

\begin{align*} x^4 + y^4 \amp = 16\\ \frac{d}{dx} x^4 + y^4 \amp = \frac{d}{dx} 16\\ 4x^3 + 4y^3 \frac{dy}{dx} \amp = 0\\ \frac{dy}{dx} \amp = \frac{-x^3}{y^3} \end{align*}

For any point \((x,y)\) on this locus, the slope of the tangent at that point is given by the expression \(\frac{-x^3}{y^3}\text{.}\)

The slope is \(0\) at the point \((0, 2)\text{.}\)
The slope is undefined at \((-2, 0)\text{.}\)
The slope is \(-1\) at \((\sqrt[4]{8}, \sqrt[4]{8})\text{.}\)
The slope is \(1\) at \((\sqrt[4]{8}, -\sqrt[4]{8})\text{.}\)

These four points and matching tangent lines are drawn in Figure 1.2.1.

In the example, the slope wasn’t defined everywhere. The slope approaches vertical near the undefined point \((\pm -2, 0)\text{;}\) this makes sense, since a vertical line has no slope. This is also the point where the assumption, that \(y\) can be expressed as a function of \(x\text{,}\) breaks down. Implicit derivatives can also fail to find slopes at places where a loci self-intersects or has a sharp corner. I will show examples of these in the Section 2.1. The key idea is this: anywhere the assumption that \(y\) is a function of \(x\) fails, the process of implicit differentiation also fails.

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