Exercise 5.3 · Q6: $\frac{dy}{dx}$ if $x^3+x^2y+xy^2+y^3=81$

Basics you need

Solution

1. Differentiate each term with respect to $x$

   $$\begin{aligned} &\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) \\ &\quad + \frac{d}{dx}(y^3) = \frac{d}{dx}(81). \end{aligned}$$
   • $\frac{d}{dx}(x^3) = 3x^2$⁠.
   • $x^2y$⁠: product rule with $u = x^2$⁠, $v = y$⁠: $$\frac{d}{dx}(x^2y) = 2x\cdot y + x^2\,\frac{dy}{dx}.$$
   • $xy^2$⁠: product rule with $u = x$⁠, $v = y^2$⁠, and the chain rule for $\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}$⁠: $$\frac{d}{dx}(xy^2) = 1\cdot y^2 + x\cdot 2y\,\frac{dy}{dx}.$$
   • $\frac{d}{dx}(y^3) = 3y^2\,\frac{dy}{dx}$ (chain rule).
   • $\frac{d}{dx}(81) = 0$⁠.

   Adding up:

   $$\begin{aligned} 3x^2 + 2xy + x^2\frac{dy}{dx} &+ y^2 + 2xy\frac{dy}{dx} \\ &+ 3y^2\frac{dy}{dx} = 0. \end{aligned}$$
   Along the curve the left side is always $81$⁠, so its rate of change with respect to $x$ is $0$⁠. In $x^2y$ and $xy^2$ both factors change as $x$ changes, which is why each gives two terms. Every time a power of $y$ is differentiated, the chain rule attaches a factor $\frac{dy}{dx}$⁠.
   Product rule (2), Section 5.3; chain rule, Theorem 4, Section 5.3.1; $(x^n)' = nx^{n-1}$⁠, Table 5.3. Basics: derivative rules

2. Collect the $\frac{dy}{dx}$ terms and factor

   Keep the three terms containing $\frac{dy}{dx}$ on the left and move the other three to the right:

   $$\begin{aligned} (x^2 + 2xy + 3y^2)\,\frac{dy}{dx} &= -3x^2 - 2xy - y^2 \\ &= -(3x^2 + 2xy + y^2). \end{aligned}$$
   The coefficients of $\frac{dy}{dx}$ are $x^2$ (from $x^2y$⁠), $2xy$ (from $xy^2$⁠) and $3y^2$ (from $y^3$⁠). The remaining terms $3x^2$⁠, $2xy$⁠, $y^2$ change sign when moved across.
   Factoring out the common unknown $\frac{dy}{dx}$⁠. Basics: solving for dy/dx

3. The divisor is never zero on the curve

   Complete the square in the bracket:

   $$\begin{aligned} x^2 + 2xy + 3y^2 &= (x^2 + 2xy + y^2) + 2y^2 \\ &= (x + y)^2 + 2y^2. \end{aligned}$$

   Both parts are squares, so the bracket is $\ge 0$⁠, and it is $0$ only if $x + y = 0$ and $y = 0$⁠, that is, only at $(0, 0)$⁠. But $(0, 0)$ is not on the curve: there the left side of the relation is $0$⁠, not $81$⁠. So at every point of the curve the bracket is positive.

   Before dividing we must know the divisor is not zero. Here it never is, so the formula below holds at every point of the curve, with no exceptions. (This also shows that, for each fixed $x$⁠, the left side of the relation is strictly increasing in $y$⁠, because its derivative with respect to $y$ is this same bracket, and it runs from $-\infty$ to $\infty$ like any cubic in $y$⁠; so each $x$ gives exactly one $y$⁠, and the whole curve is the graph of one function. This aside uses two facts from outside this chapter, named in the note below.)
   A sum of squares is zero only when each square is zero (Class IX/X algebra). For the aside: a positive derivative makes a function increasing (Theorem 1, Section 6.3, Chapter 6), and a continuous function takes every value between two of its values (the intermediate value property, not proved in the book).

4. Divide

   $$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}.$$
   Division by the bracket is allowed at every point of the curve by Step 3.

Every fact used, and where it comes from