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Section 2.7 Derivatives of Exponential Functions

Now that we understand how derivatives interact with products and quotients, we are able to compute derivatives of

  • polynomials,
  • rational functions, and
  • powers and roots of rational functions.

Notice that all of the above come from knowing  1 Differentiating powers and roots of functions is actually quite a bit easier once one knows the chain rule — which we will discuss soon. the derivative of \(x^n\) and applying linearity of derivatives and the product rule.

There is still one more “rule” that we need to complete our toolbox and that is the chain rule. However before we get there, we will add a few functions to our list of things we can differentiate  2 One reason we add these functions is that they interact very nicely with the derivative. Another reason is that they turn up in many “real world” examples.. The first of these is the exponential function.

Let \(a \gt 0\) and set \(f(x) = a^x\) — this is what is known as an exponential function. Let's see what happens when we try to compute the derivative of this function just using the definition of the derivative.

\begin{align*} \diff{f}{x} &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\\ &= \lim_{h \to 0} a^x \cdot \frac{a^{h} - 1}{h} = a^x \cdot \lim_{h \to 0} \frac{a^{h} - 1}{h} \end{align*}

Unfortunately we cannot complete this computation because we cannot evaluate the last limit directly. For the moment, let us assume this limit exists and name it

\begin{align*} C(a) &= \lim_{h \to 0} \frac{a^{h} - 1}{h} \end{align*}

It depends only on \(a\) and is completely independent of \(x\text{.}\) Using this notation (which we will quickly improve upon below), our desired derivative is now

\begin{align*} \diff{}{x}a^x &= C(a)\cdot a^x. \end{align*}

Thus the derivative of \(a^x\) is \(a^x\) multiplied by some constant — i.e. the function \(a^x\) is nearly unchanged by differentiating. If we can tune \(a\) so that \(C(a) = 1\) then the derivative would just be the original function! This turns out to be very useful.

To try finding an \(a\) that obeys \(C(a)=1\text{,}\) let us investigate how \(C(a)\) changes with \(a\text{.}\) Unfortunately (though this fact is not at all obvious) there is no way to write \(C(a)\) as a finite combination of any of the functions we have examined so far  3 To a bit more be precise, we say that a number \(q\) is algebraic if we can write \(q\) as the zero of a polynomial with integer coefficients. When \(a\) is any positive algebraic number other \(1\text{,}\) \(C(a)\) is not algebraic. A number that is not algebraic is called transcendental. The best known example of a transcendental number is \(\pi\) (which follows from the Lindemann-Weierstrass Theorem — way beyond the scope of this course).. To get started, we'll try to guess \(C(a)\text{,}\) for a few values of \(a\text{,}\) by plugging in some small values of \(h\text{.}\)

Let \(a =1\) then \(C(1) = \ds \lim_{h \to 0} \frac{1^h-1}{h} = 0\text{.}\) This is not surprising since \(1^x=1\) is constant, and so its derivative must be zero everywhere. Let \(a =2\) then \(C(2) = \ds \lim_{h \to 0} \frac{2^h-1}{h}\text{.}\) Setting \(h\) to smaller and smaller numbers gives

\(h\) 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001
\(\tfrac{2^h-1}{h}\) 0.7177 0.6956 0.6934 0.6932 0.6931 0.6931 0.6931

Similarly when \(a=3\) we get

\(h\) 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001
\(\tfrac{3^h-1}{h}\) 1.1612 1.1047 1.0992 1.0987 1.0986 1.0986 1.0986

and \(a=10\)

\(h\) 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001
\(\tfrac{10^h-1}{h}\) 2.5893 2.3293 2.3052 2.3028 2.3026 2.3026 2.3026

From this example it appears that \(C(a)\) increases as we increase \(a\text{,}\) and that \(C(a) = 1\) for some value of \(a\) between \(2\) and \(3\text{.}\)

We can learn a lot more about \(C(a)\text{,}\) and, in particular, confirm the guesses that we made in the last example, by making use of logarithms — this would be a good time for you to review them.