Basic Question: Differentiating Related Functions (Calculus)In

Question

In case anyone else is searching for an answer in the future:

The question contained an assumption. There is no rule dictating different processing of the 2 variables based on the variable whose derivative you're seeking (x and y vis a vis dx/dt and dy/dt), although lord knows one might be excused for thinking there could be an extra rule that the calculus student is expected to intuit without need for explicit instruction. The fact that most problems encountered in my textbook did almost always pair x with dx/dt is a coincidence. For xy=4 you start your calculations using the product rule because--well -- xy is a product. For 12x² = y³ - 24 you start with the power rule with 12x². The downstream consequence of applying the derivative rules to the different equations is that you use x • dy/dt in one problem and x • dx/dt in the other. But nothing was done with the intention of getting that result. By analogy to algebra, for 2x=4 you'd start using division x=4/2 on the right side of the equation but for 2³-24=-16 you'd start by evaluating the exponent 2³=8 on the left side. But there's certainly no rule that you always start processing the right side of an equation before the left. In practice, most of the problems students see have all the heavy work on the left side of the equation, but all other order-of-operations rules have precedence over the left-to-right rule.

As you can see below, one related functions solution begins with x • dy/dt and another begins with x • dx/dt.

Why?

dy/dt = dy/dx • dx/dt

y'(t)= y'(x) • x'(t)

What part of the calculations is functioning as dy/dx?

EXAMPLE 1
Find dx/dt:

xy = 4

To solve, you start with:

x • dy/dt + y • dx/dt = 0

EXAMPLE 2
Find dy/dt:

12x² = y³ - 24

In contrast to Example 1, you begin solving the problem with x • dx/dt:

24x • dx/dt = 3y² • dy/dt

COMPLETE WORKED PROBLEMS (FOR REFERENCE):

EXAMPLE 1:

Given:

xy=4

dy/dt=1

y=.5

Find dx/dt

Solution to Example 1:

x = 4/y = 4/.5 = 8

x•y=4

d/dt [ X(t) + Y(t) ] = d/dt 8

x' • y + y' • x = 0

dx/dt • y + dy/dt • x = 0

dx/dt • .5 + 1 • 8 = 0

.5 • dx/dt = -8

dx/dt = -8/.5 = -16

EXAMPLE 2:

Given:

12x² = y³ - 24

dx/dt= -6

x= -.5

Find dy/dt

Solution to Example 2:

12•(-.5)² = y³ - 24

27=y³

y=3

12x² = y³ - 24

24 • [X(t)] • dx/dt = 3 • [Y(t)]² • dy/dt

{24 • [X(t)] • dx/dt} / 3 • [Y(t)]² = dy/dt

(24 • x • dx/dt) / (3 • y²) = dy/dt

(24 • -.5 • -6) / 27 = dy/dt

dy/dt = 72/27 = 8/3

Answer 1

Think of $x$ and $y$ as functions of $t$. So in that example you would just do the product rule $$\frac{d}{dt} (xy) = x \frac{dy}{dt} + y \frac{dx}{dt} $$ or $$\frac{d}{dt} [x(t)y(t)] = x(t)y'(t) + x'(t)y(t)$$ Then for harder problems you're essentially just doing the chain rule. Like for $$\frac{d}{dt} [x^2+y^2] = \frac{d}{dt}[x(t)^2+y(t)^2] = 2xx'(t) + 2yy'(t) = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} $$ So I suppose the general rule would be to treat your variables as functions of $t$ and do normal calculus on functions.

Answer 2

Why? Because you're using two different rules about differentiation to handle different quantities.

In the first case you are finding $\frac{d}{dt}$ of $xy$, and $x\frac{dy}{dt}$ is the first part of the product rule (the second part is "$+ \frac{dx}{dt}y$").

In the second case you are finding $\frac{d}{dt}$ of $12x^2$, and, if we temporarily write $w(x) = 12x^2$, then the chain rule says $\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} = 12\cdot2\cdot x^1 \cdot \frac{dx}{dt}$, using what you know about differentiating the function $x^2$. And in this second case you are using the chain rule once again on the right hand side, to deal with $\frac{d}{dt}$ of the $y^3$.

As for your "What part...." question: you're gonna have trouble if you use that version of the chain rule, with those variables, in this situation. Let's use completely new variables to try to make things clear. You use the chain rule when $a$ depends on $b$, and $b$ depends $c$. This means that $a$ depends on $c$ (when you change $c$, $b$ changes, which then changes $a$.) When you want to find exactly how $a$ changes with $c$ (i.e. calculate $\frac{da}{dc}$), the chain rules tells you that

$$ \frac{da}{dc} = \frac{da}{db} \cdot \frac{db}{dc}$$

So when you write

$$ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$

this implies that $y$ depends on $x$, and $x$ depends on $t$. But in your situation $x$ and $y$ both depend on $t$, separately. (Of course, there is some relationship between $x$ and $y$ in both of your problems - that's the point of related rates - but we're thinking of them as separate functions here, and are curious how each changes as $t$ changes, i.e. $\frac{d}{dt}$.