Some points that satisfy the equation \(y^3-4y=x^2-1\) are graphed in Figure 7.1.1. Clearly this set of points does not constitute a function where \(y\) is a function of \(x\text{;}\) for example, there are three points that have an \(x\)-coordinate of \(1\text{.}\)

To get around this problem we are going to employ a technique called implicit differentiation. We use this technique to find the formula for \(\lz{y}{x}\) whenever the equation relating \(x\) and \(y\) is not explicitly solved for \(y\text{.}\) What we are going to do is treat \(y\) as if it were a function of \(x\) and set the derivatives of the two sides of the equation equal to one another. This is actually a reasonable thing to do because so long as we are at a point on the curve where the tangent line is not vertical, we could make \(y\) a function of \(x\) using appropriate restrictions on the domain and range.

Since we are treating \(y\) as a function of \(x\text{,}\) we need to make sure that we use the chain rule when differentiating terms like \(y^3\text{.}\) When \(u\) is a function of \(x\text{,}\) we know that \(\lzoo{x}{u^3}=3u^2\lzoo{x}{u}\text{.}\) Since the name we give \(\lzoo{x}{y}\) is \(\lz{y}{x}\text{,}\) it follows that when \(y\) is a function of \(x\text{,}\) \(\lzoo{x}{y^3}=3y^2\lz{y}{x}\text{.}\) The derivation of \(\lz{y}{x}\) for the equation \(y^3-4y=x^2-1\) is shown in Example 7.1.2.

At first it might be unsettling that the formula for \(\lz{y}{x}\) contains both the variables \(x\) and \(y\text{.}\) However, if you think it through you should conclude that the formula must include the variable \(y\text{;}\) otherwise, how could the formula generate three different slopes at the points \(\point{1}{2}\text{,}\) \(\point{1}{0}\text{,}\) and \(\point{1}{-2}\text{?}\) These slopes are given below. The reader should verify their values by drawing lines onto Figure 7.1.1 with the indicated slopes at the indicated points.