Before coming to the next lab session, Lab 8, read the special note at the beginning of that lab. You are asked to do some reading outside of class before you begin working on the lab in class. More explanation is given in that note.

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 \(x\); for example, there are three points that have an \(x\)-coordinate of \(1\).

Nevertheless, there is a unique tangent line to the curve at each point on the curve and so long as the tangent line is not vertical it has a unique slope. We still identify the value of this slope using the symbol \(\lz{y}{x}\), so it would be helpful if we had a formula for \(\lz{y}{x}\). If we could solve \(y^3-4y=x^2-1\) for \(y\), finding the formula for \(\lz{y}{x}\) would be a snap. Hopefully you quickly see that such an approach is just not reasonable.

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

Example7.1.2

\begin{align*}
y^3-4y&=x^2-1\\
\lzoo{x}{y^3-4y}&=\lzoo{x}{x^2-1}&&\text{Differentiate each side with respect to $x$.}\\
3y^2\lz{y}{x}-4\lz{y}{x}&=2x&&\text{Respect the chain rule with the $y$-terms.}\\
\left(3y^2-4\right)\lz{y}{x}&=2x&&\text{Begin to isolate $\lz{y}{x}$}\\
\lz{y}{x}&=\frac{2x}{3y^2-4}.
\end{align*}

At first it might be unsettling that the formula for \(\lz{y}{x}\) contains both the variables \(x\) and \(y\). However, if you think it through you should conclude that the formula must include the variable \(y\); otherwise, how could the formula generate three different slopes at the points \(\point{1}{2}\), \(\point{1}{0}\), and \(\point{1}{-2}\)? 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.

Use the process of implicit differentiation to find a formula for \(\lz{y}{x}\) for the curves generated by each of the following equations. Do not simplify the equations before taking the derivatives. You will need to use the product rule for differentiation in Exercises 7.1.1.4–7.1.1.6.

1

\(3x^4=-6y^5\)

2

\(\fe{\sin}{x}=\fe{\sin}{y}\)

3

\(4y^2-2y=4x^2-2x\)

4

\(x=y\,e^y\)

5

\(y=xe^y\)

6

\(xy=e^{xy-1}\)

7

Several points that satisfy the equation \(y=xe^y\) are graphed in Figure 7.1.3.

Find the slope and equation of the tangent line to this curve at the origin. (Please note that you already found the formula for \(\lz{y}{x}\) in Exercise 7.1.1.5.