In Chap. 3, we found derivatives of a few functions that were sums of other functions. Perhaps you noticed that the derivative of the whole sum was the same as the sum of the derivatives of its addends. That was no accident!
Definition and notation
Let’s be sure we know what’s meant by the sum of two functions. When we want to add two functions, we simply add the expressions that define those functions. For example, if
f1(x ) = 6x 3 and
f2(x ) = −3x then
f1(x ) +f2(x ) = 6x 3− 3x Here’s another example. If
f1(x ) = −2x2 and
f2(x ) = 5x+ 3
Sum Rule 75
then
f1(x ) +f2(x ) = −2x 2+ 5x+ 3
When two functions f1 and f2 operate on the same variable such as x, we can denote their sum in various ways. Here are the three most common formats:
f1(x)+f2(x ) ( f1+f2) (x )
f1+f2
The rule in brief
The sum rule for two derivatives tells us this:
• The derivative of the sum of two differentiable functions of a single real variable is equal to the sum of their derivatives.
This rule always works, as long we stick with the same variable in both functions, and as long as the functions are both differentiable at all the points that interest us. Stated symbolically, if f1 and f2 are differentiable functions of the same variable, then
(f1+f2)′ =f1′ +f2′
Are you confused?
Have you noticed that sometimes we write f (x ) to denote a function and f ′(x ) to denote its derivative, and then later we write f and f ′ to denote the same things? Don’t let this bother you. As long as we know thatf and f ′ operate on the same variable, it doesn’t matter whether or not we include the variable in parentheses after the name of the function.
When we want to be sure we know which variable we’re talking about, then it’s a good idea to include the parentheses and the variable. This can be important in multi-variable calculus, which we’ll study later in this book. It’s also important if we’re talking about something other than the independent variable itself, such as the number 4 or the expression (x+ Δx ).
Example
Consider the following functions of x. Note that the third function is the sum of the first two:
f1(x ) = 6x 3 f2(x ) = −3x ( f1+f2) (x ) = 6x 3− 3x
We can find the derivatives of the first two functions using rules we’ve already learned:
f1′(x ) = 18x 2 f2′(x ) = −3 The sum of the derivatives is
f1′(x ) +f2′(x ) = 18x 2− 3
If we use the definition at the end of Chap. 3 to calculate the derivative of the sum function using limits, we will get
( f1+f2)′ (x ) = 18x 2− 3
which is the same as the result we get when we find the derivatives of component functions separately and then add them.
Another example
Let’s look at an example involving functions of the variable y. Again, the third function is the sum of the first two:
f1(y ) = −2y2 f2(y ) = 5y+ 3 ( f1+f2) ( y ) = −2y2+ 5y+ 3
When we find the derivatives of these two component functions, we get f1′(y ) = −4y
f2′(y ) = 5 The sum of these derivative functions is
f1′(y ) +f2′(y ) = −4y+ 5
Again using the definition at the end of Chap. 3, we can show that ( f1+f2)′ ( y ) = −4y+ 5
This is what we get when we determine the derivatives separately and add them.
Mathematical induction
If you’ve taken any number-theory or set-theory courses, you’ve learned about a technique calledmathematical induction. If you haven’t seen this technique before, or if you’ve forgot-ten how it works, here’s a brief review. Mathematical induction allows you to prove infinitely many facts in a finite number of steps.
Sum Rule 77
Imagine an infinite sequence of statements called S1, S2, S3, and so on. Suppose that you want to prove all these statements true. You can’t prove every statement one by one, because you don’t have an infinite amount of time. But suppose that you can demonstrate two things:
• The first statement S1 is true, and
• If a statement Sn (where n can be any positive integer) in the sequence is true, then the next statement Sn+1 in the sequence is also true.
Once you have done these two things, the first statement “automatically” proves the second statement. That, in turn, “automatically” proves the third statement. The logical process can go on forever, like an endless chain reaction! It proves that all the statements in the sequence are true, even if there are infinitely many of them.
Are you confused?
By now you might wonder how mathematical induction can help you prove anything that has to do with adding derivatives. The following “challenge” will answer that question!
Here’s a challenge!
Extrapolate the sum rule for two derivatives to show that it works for the sum of any number of derivatives.
Solution
The sum rule for two derivatives says that for any two differentiable functions f1 and f2, ( f1+f2)′ =f1′ +f2′
Let’s call this statement S1 in an infinite sequence of truths we intend to prove. We are given the fact that this initial statement is true. Now imagine an unlimited supply of differentiable functions called f1,f2,f3, . . . , and so on, forever! According to our sequencing scheme, the second statement, S2 (which we haven’t proven yet) is
( f1+f2+f3)′ =f1′ +f2′ +f3′ The third statement, S3 (which, again, remains to be proved) is
( f1+f2+f3+f4)′ =f1′ +f2′ +f3′ +f4′
Imagine that we have been assured of the truth of statement Sn saying that, for any n+ 1 differentiable functionsf1,f2,f3, . . . , and so on up to fn+1,
( f1+f2+f3+ · · · +fn+1)′ =f1′ +f2′ +f3′ + · · · fn+1′
We must show that if Sn is true, then the next statement Sn+1 is also true, which would tell us that for any n+ 2 differentiable functions f1,f2,f3, . . . , and so on up to fn+2,
( f1+f2+f3+ · · · +fn+1+ fn+2)′ =f1′ +f2′ +f3′ + · · · +fn+1′ +fn+2′ Now let’s temporarily change a couple of names, like this:
• Call the sum of functions f1+f2+f3+ · · · +fn+1 by the nickname g
• Call the sum of functions f1′ +f2′ +f3′ + · · · +fn+1′ by the nickname h On that basis, we can rewrite Sn as
g′ =h
Let’s pick a function “out of the air” and call it fn+2. (We can call it anything we want, so why not that?) We add this new function to g and then take the derivative:
(g+fn+2)′
According to the sum rule for two derivatives, we know that (g+fn+2)′ =g′ +fn+2′
We also know that g′ =h, so we can substitute in the above equation to get (g+fn+2)′ =h+fn+2′
When we give g and h their original names back, the above expression expands to ( f1+f2+f3+ · · · +fn+1+fn+2)′ =f1′ +f2′ +f3′ + · · · +fn+1′ +fn+2′
That’s exactly the statement Sn+1! Having fulfilled the requirements of the mathematical induction routine, we’re entitled to claim that the derivative of a sum is equal to the sum of the derivatives, no matter how many functions are involved. We can now call this rule simply the sum rule for derivatives.