## Derivatives of exponentials summary

## The derivative of $e^x$

The derivative of $e^x$ is

$$\frac{d}{dx} \Big( e^x \Big) = e^x.$$

We can prove this by going back to the definition.

$$\begin{eqnarray}

\frac{d}{dx} \Big( e^x \Big) &= \lim_{h\to 0} \frac{e^{x+h}-e^x}{h}\\

& = \lim_{h\to 0} \frac{e^{x}e^h-e^x}{h}\\

&= e^x \lim_{h\to 0} \frac{e^{h}-1}{h}.

\end{eqnarray}$$

To finish the derivation we need to a special limit that you may (or may not) have covered in your class. The limit we need is $$\lim_{h\to 0} \frac{e^h-1}{h} = 1.$$

Using this limit we can see that indeed

$$\frac{d}{dx} \Big( e^x \Big) = e^x.$$

Let’s use this new derivative in a few examples to get comfortable with it.

#### Example

Let $h(x) = x e^x$ what is $h'(x)?$

#### Solution

We begin by noticing that $h(x)$ is a product of the functions $x$ and $e^x$. Therefore we can apply the product rule to get $$h'(x) = 1 \cdot e^x + x \cdot e^x = e^x + xe^x.$$

#### Example

Let $f(x) = \left( e^x\right)^2$ what is $f'(x)?$

#### Solution

There are actually two ways of doing this problem, both of which involve the chain rule. Let’s take a look at them both. For the first we begin by looking at the function as it was given to us and notice that this is a function composition with inner function $g(x) = e^x$ and outer function $h(x) = x^2$, therefore when we apply the chain rule we get

$$\begin{eqnarray}

f'(x) & = h'(g(x))g'(x)\\

& = 2g(x)\cdot g'(x)\\

& = 2 e^x e^x = 2e^{2x}.

\end{eqnarray}$$

Alternatively we can use exponent rules to rewrite

$$f(x) = \left( e^x)\right)^2 = e^{2x}.$$

In this case we have an inner function of $g(x) = 2x$ and an outer function of $h(x) = e^x$. Applying the chain rule we get:

get

$$\begin{eqnarray}

f'(x) & = h'(g(x))g'(x)\\

& = e^{g(x)}\cdot g'(x)\\

& = e^{2x}\cdot 2 = 2e^{2x}.

\end{eqnarray}$$

Now you try it!

## Practice Problems for derivatives exponentials

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## Derivative of $a^x$

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

This can be computed from the derivative of $e^x$ we just computed by using logarithm rules and the chain rule.

We do this by first rewriting $f(x) =a^x$ in the following way:

$$f(x) = a^x = e^{\ln (a^x)} = e^{x\, \ln a}.$$

Now what we have is a function composition where the inner function is just $x$ multiplied with a constant (I know $\ln a$ might look serious, but it is just a constant). Then using the chain rule with inner function of $g(x) = x\, \ln a$ and an outer function of $h(x) = e^x$. Applying the chain rule we get:

get

$$\begin{eqnarray}

f'(x) & = h'(g(x))g'(x)\\

&= e^{g(x)}\cdot g'(x)\\

&= e^{x\, \ln a} \ln a\\

& = a^x \ln a.

\end{eqnarray}$$

#### Example

Let $g(t) = t^2 \cdot 2^t$, what is $g'(t)$?

#### Solution

We can start by noticing that $g(t)$ is a product, meaning we need to use the product rule to take the derivative. Doing this we get

$$begin{eqnarray}g'(t) & = &\frac{d}{dt} \left( t^2 \right) 2^t+ t^2 \cdot \frac{d}{dt} \left(2^t \right)\\ &=& 2t\cdot 2^t + t^2 \cdot 2^t \ln 2.

end{eqnarray}$$

## Why is e so special?

You might be wondering why we started with $e^x$ and then extended this to $a^x$. You might even be wondering why we care about $e^x$ at all? Who decided this function was important and why do we use it all the time?

The exponential function $\exp(x) = e^x$ has quite a few different characterizations, but the one we will use is the following:

#### Definition: Exponential Functions

We can define the exponential function with base $a$ to be the continuous function $a^x$ that is defined at integers $n$ by $a^n=a\cdots a$ (the product of $n$ $a$’s), and at rational number $m/n$ as $a^{m/n} = \sqrt{m}{n}$. In other words $a^x$ should behave exactly as expected for the numbers where we can explicitly compute it using power rules.

Let’s look at the graphs of a couple of examples.

Here we can see that at $x=1$ each function has a $y$ value equal to the base of the exponential. Moreover at $x=1/2$ we can see that $4^{1/2}= \sqrt{4} =2$ and similarly with $(1/4)^x$.

#### Definition: $e^x$

The exponential function $exp(x) =e^x$ is the unique exponential function where the base $e$ satisfies

$$\lim_{h\to 0} \frac{e^h-1}{h} = 1.$$

What does this mean for us? This means that the function $e^x$ is the unique exponential function that gives itself back as the derivative. This is extremely useful in mathematical modeling and other applications. Below we can look at the graphs of $2^x, e^x, 3^x$ and their derivatives.

As we can see from the the function $f(x) = e^x$ is quite special and will end up being something we see a lot of in Calculus. Let’s finish it off with a few questions about higher derivatives of exponential functions.

## Higher derivatives of exponential functions

#### Example

Let $f(x) = e^x$ what is $f^{(10)}(x)?$

#### Solution

Since $\frac{d}{dx} (e^x) = e^x$ it doesn’t matter how many times we differentiate, we still get $e^x$ back. This means that

$$f^{(10)}(x) = e^x!$$

#### Example

Let $f(x) = 2^x$ what is $f^{(8)}(x)?$

#### Solution

Recall that $\frac{d}{dx} (2^x) = 2^x \ln 2.$ This means that for each time we differentiate we should get another multiple of $\ln 2$. For example:

$$f”(x) = \frac{d}{dx} \Big(2^x \ln 2 \Big) = \ln 2 \frac{d}{dx} \Big(2^x \Big) = (\ln 2)^2 2^x.$$

This means that if we differentiate $8$ times we get:

$$f^{(8)}(x) = (\ln 2)^8 2^x!$$