Worksheet for Section 3.2

Secton 3.2 is about derivatives of exponential functions. You saw some useful limits involving exponential functions in earlier sections, but we just explored them using tables of values -- it is time now to be more precise in order to find a general rule for derivatives of exponential functions. Starting with the function $f(x)=2^x$, compute the derivative from the limit definition:

$$f'(x)=\lim_{h\to 0}\frac{2^{x+h}-2^x}{h}=2^x\cdot\lim_{h\to 0}\frac{2^h-1}{h}$$

Look at this remaining limit, $$\lim_{h\to 0}\frac{2^h-1}{h}$$, using the Limits applet. What value do you get? This same procedure can be used to explore the general exponential function $g(x)=a^x$:

$$g'(x)=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}=a^x\cdot\lim_{h\to 0}\frac{a^h-1}{h}$$

Now explore the limit $$\lim_{h\to 0}\frac{a^h-1}{h}$$ in the Limits applet, using several specific values of $a$ (e.g. $a=2,3,4,5,...$). In each case, the limit is a constant. This suggests a derivative rule of the following form:

$$\frac{d}{dx}a^x=k\cdot a^x$$

where the value of the constant $k$ depends on the value of the base $a$. This raises a new question, though: is there a value of $a$ for which $k=1$? Look again at the limits you explored above with $a=2$, $a-3$, etc. -- does it look like there should be such a value?

Recall that the constant $k$ comes from the limit $$\lim_{h\to 0}\frac{a^h-1}{h}$$ -- so you are really looking for a value of $a$ for which this limit is 1. For small values of $h$, then, $$\frac{a^h-1}{h}\approx 1$$. Solving for $a$: $a\approx (1+h)^{1/h}$. But now explore this expression for small values of $h$ using the Limits applet -- it turns out that $\lim_{h\to 0}(1+h)^{1/h}=e$. From this, it can further be shown that $$\lim_{h\to 0}\frac{e^h-1}{h}=1$$, as desired, which gives the following derivative rule:

$$\frac{d}{dx}e^x=e^x$$

Now return to the general exponential function $g(x)=a^x$. With a bit of rewriting, this can also be computed using the limits above involving $e$:

$$g'(x)=a^x\cdot\lim_{h\to 0}\frac{a^h-1}{h} =a^x\cdot \lim_{h\to 0}\frac{e^{(\ln a)h}-1}{h}$$

Now make the substitution $t=(\ln a)h$, and notice that as $h\to 0$, so also $t\to 0$. This leads to the derivative rule:

$$\frac{d}{dx}a^x=(\ln a)a^x$$

Keep in mind in particular that this is different from the Power Rule in the last section. Here are some examples that illustrate the difference -- in each case, compute the derivative using whichever rule applies:

$$f(x)=e^\pi\qquad g(x)=x^\pi\qquad h(x)=\pi^x$$

Combining these rules with the rules from the previous section expands the range of functions for which derivatives can now be computed. Here are a few examples -- again, compute the derivative of each function using the rules you have so far:

$$f(x)=2x^3-3e^x\qquad g(x)=3\cdot 4^x + 4\cdot x^3\qquad h(x)=\sqrt{x}-x^{-2}$$