In this section, we'll learn how to find the derivatives of exponential and logarithmic functions. To begin that, we'll take a closer look at a general exponential function, armed with our new view of mathematics that includes limits. ... but first, let's brush up on the laws of exponents:

Product |
a^{m} ·a^{n} = a^{m+n} |
2^{2} · 2^{3} = (2·2)(2·2·2) = 2^{5} |

Quotient |
||

Power |
(a^{m})^{n} = a^{m·n} |
(2^{2})^{3} = (2·2)(2·2)(2·2) = 2^{6} |

Inverse |
||

Zero power |
a^{0} = 1 |
Why? We need a^{m}a^{n} = a^{m·n} when m = 0. In order for this law to be satisfied when m = 0, we have a^{n} = a^{m+n} = a^{0+n} = a^{0}a^{n}, so a^{0} must be 1. |

Let's consider the function **f(x) = a ^{x}** in a little more detail. We know it's defined for x ∈ {

We also know that the function is defined for x ∈ {**Q**}, the rational numbers, because we know that

,

which is exactly defined. But what about something like

How do we calculate the value of a number raised to an **irrational** power? To wrap our heads around this dilemma, let's replace **a** with a number:

Then let's approximate the root of three using rational numbers, one on the low side and another on the high side:

Note that 1.7 and 1.8 are rational numbers. For example, 1.7 is 17/10. If √3 can be sandwiched between two rational numbers, then

The value of 2^{√3} must lie between 2^{1.7} and 2^{1.8} somewhere. Now the trick is that we can do arbitrarily better than that. We can improve our rational-number estimates of 2^{√3}, getting arbitrarily close to it.

There are two consequences to all of this. First, any **graph** of such an exponential function really has to be considered to be "full of **holes**", as depicted in the graph below. There are an infinity of holes between each pair of irrational exponents.

Second, it means that the value of an exponential function is really a **limit** at those points:

Notice that this is the very definition of the continuity of a function, and exponential functions are continuous.

Now to find the derivative of the general exponential function **f(x) = a ^{x}**. We begin with the difference quotient, as usual. In the third step, we apply the product property of exponents.

Next we factor out the common **a ^{x}** and realize that it plays no factor in evaluating the limit (doesn't contain h), so it can be moved to the front.

We see then that the derivative of an exponential function is just the function itself times some value, the value of the limit:

,

which we assume to take on some constant value. That idea is worth repeating because it's quite an important one:

The derivative of an exponential function is directly proportional to the function itself:

Now let's think a little harder about that limit. In the upper table below, I've calculated some of its values with different bases (0.5, 1, 2, 3) and with decreasing h (i.e. as h→0). In each case the limit converges to some number (bottom row), but there's an interesting possibility ...

If we choose **a** to lie somewhere between 2 and 3, it should be possible (and it is) to get the limit to converge to **one**. In that case, the derivative of the exponential function would just be the exponential function itself.

The lower table ( ↑ ) shows that the number 2.7182818 ..., which turns out to the the irrational number ** e**, the base of all continuously-growing exponential functions. Use of

The natural exponential function is its own derivative:

The natural exponential function, **f(x) = e ^{x}** is the only function that is its own derivative. That means that the value,

Now we just need an expression for the derivative of **f(x) = a ^{x}**, where

Now **ln(a)** is a constant, so it's just a matter of applying the chain rule to take the derivative, like this:

Remember that **e ^{(ln a)} = a**, then we can use that to find the derivative rule for a general exponential function:

The derivative of an exponential function with a base other than **e** is

Notice that this equation works if the base is **e**, too, because **ln(e) = 1**.

Solution: This is a product-rule problem that contains an exponential part; the function is a product of the quadratic function **4x ^{2}** (which is

The derivative is

which simplifies a bit to

Solution: This is a composite function, an exponential function f(a) = 4^{a} where a = 2x. The derivative is

The derivative of 2x is 2, so the derivative of the function is:

You'll need to memorize these two exponential derivatives, but a big part of finding derivatives of these functions is applying the other rules of differentiation.

Solution:

Taking the inner derivative gives

Once we know the derivative of **e ^{x}**, the derivative of its inverse function,

First use exponentiation with the base ** e** to get rid of the log, a common manipulation with log equations,

Now take the derivative of each side, remembering to use the chain rule on **e ^{y}**, because

and

Then the trick is to realize that in the second step, labeled (*), we set **e ^{y} = x**. That allows us to solve for the derivative of

So our derivative is

The derivative of the natural exponential function f(x) = ln(x) is

All that's left for us to do in this section is to generalize the derivative of **ln(x)** so that we have a method of differentiating *any* log function (with any base). To do this, we simply recall that to change the base of a base * b* logarithm to the base

From there, if we recognize that **1/ln(b)** is just a constant, we can use the previous result to get

And that's our result for the derivative of a log function of any base.

The derivative of a general (any base) logarithmic function is

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.