Print preview
Handout Monday 9/29
We know how to take derivatives of a bunch of basic functions, as well as how to take derivatives of functions that are the sum or difference of those functions. The sum rule says that the derivative of a sum is just the sum of the derivatives. That is, if \(f(x)\) is the sum of two functions, we can find \(f'(x)\) by taking the derivatives of each of the two functions and adding the results.
However, this is not true for products. On the one hand, we can quickly see that this fails if we take the derivative of \(f(x) = x^3\cdot x^7\text{.}\) On one hand, we could simplify this function as \(f(x) = x^{10}\text{,}\) so \(f'(x) = 10x^{9}\text{.}\) However, if we took the derivative of the two parts separately, we would have \(3x^2\) and \(7x^6\text{.}\) Multiplying those would give \(21x^8\) which is definitely not the same.
Another reason to be skeptical: suppose a magic rectangle has sides \(l\) and \(w\text{,}\) both of which are increasing with time. The area of the rectangle is then also a function of time. That is, \(A(t) = l(t)\cdot w(t)\text{.}\)
-
Let’s say that \(l(5) = 20\) and \(w(5) = 30\text{.}\) Then \(A(5) = 600\text{.}\) How much new area do you get if \(l(t)\) increases by \(3\) and \(w(t)\) increases by \(4\) in the next day? Does the area just increase by \(12\text{?}\)
-
No, since \(l(6)\) growing to 23 doesn’t just add 3 feet: we get area of \(3 \cdot 30 = 90\) more feet! And when the width increases by 4 feet, that adds \(4 \cdot 20 = 80\) more feet.
-
Draw the picture. Every increase in one dimension must be multiplied by the current size of the other dimension.
-
In other words, the increase is \(\Delta l \cdot w + \Delta w \cdot l\text{.}\)
This gives us the product rule: if \(P(x) = f(x)\cdot g(x)\text{,}\) then \(P'(x) = f'(x)g(x) + f(x) g'(x)\text{.}\)
Do a few examples with functions we know. Start with \(x^3x^7\text{.}\)
Quotients have a similar rule: the quotient rule says that if \(Q(x) = \frac{f(x)}{g(x)}\) then
\begin{equation*}
Q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\text{.}
\end{equation*}
We can “prove” this from the product rule by noting that \(f(x) = Q(x)g(x)\text{.}\) Apply the product rule and then solve for \(Q'(x)\text{.}\)
Try some examples.
Note that we can use both rules in the same function. A function that is the ratio of a product and another function, for example.
