MAT1193 – 5c Sum, Product and Quotient Rules In this set of notes, we will consider the rules for applying derivatives to three basic ways for combining functions. Remember that there are three perspectives that we can take on each of our rule for the derivative: 1. The derivation of the rule; 2. An intuition for the rule; 3. The application of the rule. We will try to get some insight (1 & 2) into the sum and product rules, but not for the quotient rule. To start, lets just state the rules. I will use both the prime notation and the differential notation. Sum rule - the derivative of the sum of two functions is the sum of the derivative of each function: h(x) = f (x) + g(x) →h# ( x) = f #( x) + g# ( x) dh df dg h(x) = f (x) + g(x) → = + dx dx dx Here are some examples:

b(z) = z3 + z2 →b# ( z) = 3z2 + 2z € 3/2 −1 dp d q d q 3 3 p(q) = q3/2 + q−1 → = ( ) + ( ) = q1/2 + (−1)q−2 = q1/2 − q−2 dq dq dq 2 2 € Constant sum rule - the derivative of the sum of a function and a constant value is equal to the derivative of the function alone: h(x) = f (x) + a →h #( x) = f #( x) dh df h(x) = f (x) + a → = dx dx Examples:

f (x) = z3 + 345 → f #( x) = 3x 2 € da a(l) = l−7 − 0.17 → = −7l−8 dl € This is actually a special case of the sum rule, since we can view the constant as g(x) = a resulting from the constant function and just plug in to the sum rule. € Note that we could also make a “difference rule” that is the same as the sum rule but with a minus sign. But this rule is so natural and can be derived using a -1 in the constant product rule (explained below) that we don’t give it it’s own rule.€ The product rule - the derivative of the product of two functions is the derivative of the first function times the value of the second plus the value of the first function times the derivative of the second: h(x) = f (x)g(x) →h# ( x) = f #( x)g(x) + f (x)g# ( x) dh df dg h(x) = f (x)g(x) → = g(x) + f (x) dx dx dx It is NOT true that the derivative of the product of two functions is the product of the derivative of the two functions! € Examples: The quotient rule - the derivative of the quotient of two functions is the derivative of the numerator times the value of the second minus the value of the numerator times the derivative of the denominator, all over the denominator squared: f (x) f #( x)g(x) − f (x)g# ( x) h(x) = →h# ( x) = 2 g(x) (g(x)) df dg g(x) − f (x) f (x) dh dx dx h(x) = → = 2 g(x) dx (g(x))

Examples: This rule gets pretty complicated although the numerator of the final answer is € pretty close to the product rule.

How about applying multiple rules in one problem? For example, suppose we have (x 4 + x) the function f (x) = 2 The last operation that was done in putting this x . function together was to divide the function n(x) = x 4 + x d(x) = x 2 by the function (I just made up the names for the function n(x) and d(x)). By the quotient rule, € n (x)d(x) − n(x)d (x) " " € € f "( x) = 2 . But to apply that we need to know n’(x) and (d(x)) d’(x). Well d’(x) = 1 by the power rule. To find n’(x) we need to apply the sum rule to get n’(x) = 4x3+1. Now we can substitute these back into the formula to get

€ (4x 3 +1)x 2 − (x 4 + x)2x f "( x) = 2 . The rest is just algebra. (x 2 )

In doing these complicated problems, the differential notation is really convenient. € Combined with parentheses, it allows us to have a notation for “take the derivative of” without introducing a new name. If we redo the above problem in differential notation we can write

d x 4 x d x 2 ( + ) 2 4 ( ) x − x + x 3 2 4 df dx ( ) dx (4x +1)x − (x + x)2x = 2 = 4 dx (x 2) x With this notation and the ability to diagram functions that we learned earlier in the course, we can find the derivative of very complicated functions by applying rules of the derivative one step at a time. € Intuitions The easiest of the rules is the sum rule. It’s simple to implement since we just apply the derivative in the same form as the original function. To see that it makes sense, let’s examine the rule in terms of an application. Suppose there are two dominant species of frog in a pond in the tropics and both are sensitive to the average temperature over the winter (in oC). In particular, suppose the number of frogs the larger species depends on temperature T according to the rule l(T) = T2, while how the smaller one depends on temperature is given by the function s(T) = 1000*T-1. If the average temperature is 10oC, what is the total number of frogs? How sensitive is the population of frogs to changes in temperature, or in other words what is the rate of change in the frog population relative to temperature? In particular, use the derivative to estimate the change in the population of frogs if temperature decreases by 0.1 oC. ANS: First, let f(T) be the total number of frogs at temperature T. Since this is just the number of larger frogs plus the number of smaller frogs,

2 −1 f (T) = l(T) + s(T) = T +1000T

If we plug in T = 10, we see that there are l(10) = 102=100 larger frogs and 1000*(10)-1 = 1000/10 = 100 smaller frogs. Let’s determine the rate of change in each population. Using the power rule l’(T) = 2T. At the base point T=10, l’(10)=20 € so the population of larger frogs increases by 20 frogs for every degree increase in temperature. For the smaller frogs we have s’(T) = 1000 d(T-1)/dT by the constant product rule and s’(T) = 1000*(-1)*T-2 = -1000/T2 by the power rule. At the base point T=10, s’(10)=-1000/100=-10 so the population of smaller frogs decreases by 10 frogs for every degree increase in temperature. The sum rule just says that the rate of change in the total frog population is the rate of change in the larger frog population plus the rate of change in the smaller frog population: f’(10) = 20+(-10) = 10.

Let’s use the derivative to estimate changes in the frog population if the temperature decreases by 0.1 oC . The easiest way to see what happens is to use the differential notation and the fact that for small changes in temperature ΔT Δf df ≈ ΔT dT Then the change in frog population can be determined by

df 10 frogs o Δf ≈ ΔT = *(−0.1 C) = −1frog € dT o C In other words we expect there to be one less frog if the temperature decreases by 0.1 oC . What happens to each subpopulation? € dl 20 frogs Δl ≈ ΔT = *(−0.1oC) = −2 frog dT o C ds −10 frogs Δs ≈ ΔT = *(−0.1o C) = +1frog dT oC So the decrease in temperature should lead to 2 fewer larger frogs but one more smaller frog: the total change in the frog population is just the sum of the change in the population of large frogs plus the change in the population of the smaller frogs. € That is just the sum rule. What about the product rule? Let’s consider another example. Suppose Axel Fromkin runs a factory that makes widgets. Let w be the number of widgets produced per week, p be the price per widget and g be the factories gross income. Then g = w*p. Currently, the factory churns out 1000 widgets per week and he can sell the widgets at $10 per widget. That means of course that the factor has gross income of 1000widgets 10dollars 10,000dollars g = w * p = = week widget week

Suppose something happened that increased the number of widgets per week by 20. So the change is Δw = 20 widgets/week. What is Δg, the change in gross income? Well if he makes 20 extra widgets per week and each widget brings in 10 dollars € then Δg = Δw*p = (20 widgets/week)*(10 dollars/widget) = 200 dollars/week. After things return to normal (w = 1000 widgets/week), suppose something affects the price so that it goes down to 9 dollars/widget. The change in price Δp = -1 dollars/widget. What is Δg, the change in gross income for this situation? Axel is getting 1 dollar less per widget and so the change in income is given by Δg = w*Δp = (1000 widgets/week)*(-1 dollars/widget) = -1000 dollars/week. In other words, he’s making 1000 fewer dollars per week.

Now let’s assume that both changes happen. What is the change in gross income? One guess is that the increase in production would give Axel an extra 200 dollars per week and the change in price would give him 1000 fewer dollars per week so that the total change in gross income is

Δg = Δw*p + w*Δp = 200-1000 = -800 dollars/week Notice that this is very similar to the product rule.

In fact this answer is off by a little bit. To see that, lets let w0=1000 widgets/week and p0=10 dollars/widget be the baseline values. After the changes take place, the new production w is the baseline production plus the change in production: w = w0+Δw = 1000+20 = 1020 widgets/week Similarly we can write the new price as p = p0+Δp = 10+(-1) = 9 dollars/widgets Then the new gross income g = w*p = 1020*9 = 9180 dollars per week. But calculating all the numbers hides how the change in g depends on the change in w and the change in p. So let’s rewrite things and then multiply it out g = (w0+Δw )*(p0+Δp) = w0*p0 +Δw *p0+w0*Δp+Δw*Δp =10000+200-1000-20=9180 dollars/week

Since the baseline value for g is g0=w0*p0 =10000 dollars/week, the change in g is

Δg = Δw*p0+w0*Δp+Δw*Δp = 200-1000-20 = -820 dollars/week. This is only 20 dollars/week different than our original guess of -800 dollars/week, with that 20 dollars/week coming from the multiplication of Δw*Δp. So what would happen if the changes in production and price we one tenth as big? Then Δw = 2 and Δp = -0.1. Then

Δg = Δw*p0+w0*Δp+Δw*Δp = 2*10+1000*(-.1)+2*(-.1) =20-100-.2 dollars/week

Now our estimate based on Δw*p0+w0*Δp gives a change of -80 dollars/week and the extra bit added by Δw*Δp is only -.2 dollars/week or 20 cents/week. We can see as the changes in production and price get small we can write

Δg≈ Δw*p0+w0*Δp This is where the product rule comes from: “The change in the first variable times the value of the second plus the value of the first variable times the change in the second.”

To finish this example off, lets suppose the production and the price both depend on the cost of the raw material x, where x is measured in dollars. Lets suppose the exact formulas are given by

w(x) = 900+x2, p(x) = 20-x, and so g(x) = w(x)*p(x) = (900+x2) *(20-x). When x=10 dollars, we have w(10)=900+100=1000 widgets/week and p(x) = 20-10 = 10 dollars/widget. g(10) = w(10)*p(10) = 10000 dollars/week. Axel wants to know the rate of change in his gross income relative to the change in the price of x. Applying the product rule dg dw dp = p(x) + w(x) dx dx dx dg = (2x)(20 − x) + (900 + x 2 )(−1) dx To figure this out we needed the sum and power rule to find that w’(x) = 2x and p(x)=-1. At x=10 we have that dg/dx = 20*10+1000*(-1) = -800. What are the units of this rate of change? We want to know the rate of change of the gross income per € change in price of substance x so dg/dx has units of (dollars/week) per dollar. That is to say for every dollar increase in substance x, Axel is expected to decrease his income by 800 dollars/week. Finally, it will be the case that some student is seduced into writing g’(x) = w’(x)*p’(x) instead of properly applying the product rule to get g’(x) = w’(x)*p(x) +w(x)*p’(x). Besides being plain wrong, we can see that g’(x) = w’(x)*p’(x) is wrong by looking at the units. w’(x) is the rate of change in production with respect to x and will have units of (widgets/week) per dollar. p’(x) is the rate of change in price with respect to x and will have units of (dollars/widget) per dollar. Then w’(x)*p’(x) will have units of (dollars/week) per dollar2. That is there is an extra “per dollar.”