Expert: Paul Klarreich - 1/17/2007

Question
right out of our beautiful grade 12 calculus course..
my question is - for functions like x^2, etc. where the variable is the base, we can use chain rule to differentiate. So why then, if the variable is in the exponent can chain rule not also be used? I mean, why does the derivative of 2^x not equal x(2^(x-1))?  Instead it = 2^x*ln2.  Is there some element of the chain rule in there that I'm missing, or is it just a rule that everyone has to know, that exponential functions equal themselves * ln of the base * derivative of the exponent. Some kind of explanation would be AWEsome. thanks a lot.

Answer
Questioner:   Andrea
Category:  Calculus
 
Subject:  exponential functions
Question:  right out of our beautiful grade 12 calculus course..
my question is - for functions like x^2, etc. where the variable is the base, we can use chain rule to differentiate. So why then, if the variable is in the exponent can chain rule not also be used? I mean, why does the derivative of 2^x not equal x(2^(x-1))?  Instead it = 2^x*ln2.  Is there some element of the chain rule in there that I'm missing, or is it just a rule that everyone has to know, that exponential functions equal themselves * ln of the base * derivative of the exponent. Some kind of explanation would be AWEsome. thanks a lot.
....................................................
Hi, Andrea,

A function like  

y = x^2,

which is a (simple example of a) polynomial, is a completely different animal from:

y = 2^x,

which is an exponential function.

Why can't you use one rule for the other?  The simple answer is that your rule:

D(x^n) = n x^n-1

took a lot of work to derive, and you will, no doubt, recall the steps.  You would have to recall the steps if anyone asked you:

"How come the x^n rule above works for all values of n, including negative numbers, and including fractions like  p/q, so we can do roots?"

You would have to say:

A. We derived the special case of the rule for  n being a positive integer this way:

           (x + h)^n - x^n
f'(x) = lim ----------------
      h->0        h

and expanded using the binomial theorem and obtained the limit. It came out to nx^n-1.

B. We derived the quotient rule, then applied it to something like

f(x) = x^-k,  using  n = -k.  

We wrote that  f(x) = 1/x^k, then used the quotient rule.  After all the smoke cleared we found that the x^n rule was still good.

C. We learned about implicit differentiation and the chain rule.  Then we could take  n = 1/k, which means the k-th root of x.  We could write:

y = x^1/k, then  y^k = x, use implicit differentiation, and again, when the smoke clears, the  x^n rule still stands.

D. We can do it for  n = p/q using the chain rule again.  The  x^n rule keeps on working.  (Sounds like a commercial.)

E. What about something like  x^sqrt(2).  That's not a rational exponent, but we can use certain continuity arguments to extend it.

So we did a lot of work to prove the  x^n rule applies to many different kinds of exponents.  BUT EVERY ONE OF THEM WAS A CONSTANT.  At no time did we ever apply the x^n rule to a case where the exponent was a variable.

So the simple answer to "Why can't we use the x^n rule on 2^x" is: "We never proved that we could, and we don't go around applying some rule unless we know it applies."

......................................
So how DO we get something for the derivative of  2^x?  The following is an outline of what you will see in your calculus text:

A. We start with this limit:

lim  (1 + h)^(1/h) = e.
h->0

B. We find the derivative of  e^x in this way:
           e^(x + h) - e^x
f'(x) = lim ---------------
      h->0        h

and find a way to compute the limit (the derivative) using the limit in A.  You find that if  f(x) = e^x, then  f'(x) = e^x.  [The amazing indestructible function.]

C. For other exponential functions, such as  2^x, you write  2 = e^(ln 2)  and use other derivative and exponent properties to work out the answer.

I hope that by working this out, you will begin to see the real beauty of the 12th grade calculus course you are experiencing.