# Derivatives of Exponential Functions

In order to differentiate the exponential function

$f(x) = a^x,$

we cannot use power rule as we require the exponent to be a fixed number and the base to be a variable. Instead, we're going to have to start with the definition of the derivative:

$\begin{aligned} f'(x) &= \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{a^{x + h} - a^x}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{a^x a^h - a^x}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{a^x \big(a^h - 1\big)}{h}. \end{aligned}$

Note that $a^x$ is not affected by the limit since it doesn't have any $h$'s in it, so it is constant as far as we're concerned. We can therefore factor this out of the limit. This gives

$f'(x) = a^x \displaystyle \lim_{h \rightarrow 0} \dfrac{a^h -1}{h}.$

Now, notice that the limit we've got above is exactly the definition of the derivative of $f(x) = a^x$ at $x = 0$, i.e. $f'(0)$. Therefore, the derivative becomes

$f'(x) = f'(0) a^x.$

Note that one of the definitions of $e$ is the fact that it is the only positive number for which $\lim_{h \rightarrow 0} \frac{e^h - 1}{h} = 1$. This is exactly what we want.

Provided that we are using the natural exponent, we get the following:

$f(x) = e^x \implies f'(x) = e^x.$

For all $a$, however, we must first convert it to

$a = e^{\ln{a}}.$

Now, we can do the following:

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

From our previous result of $f'(x)$ $($also remembering that $a$ is constant, which makes $\ln{a}$ constant$),$ we get

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

To sum up, $\boxed{\frac{d}{dx} (e^x) = e^x}$ and $\boxed{\frac{d}{dx} (a^x) = a^x \ln a}$.

Differentiate $e^{2x}$.

Using the chain rule with $f(x) = e^x$ and $g(x) = 2x$, we get $f \circ g(x) = f ( 2x) = e^{2x}$. We calculate that $f'(x) = e^x$ and $g'(x) = 2$. Thus,

$(f \circ g)' (x) = \big( f' \circ g (x) \big) \times g'(x) = f' (2x) \times 2 = 2 e^{2x}.$

Hence, the derivative of $e^{2x}$ is $2 e^{2x}$. $_\square$

Differentiate $2 ^ x$.

We first convert into base $e$ as follows:

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

Next, we apply the chain rule with $f(x) = e^x$ and $g(x) = x \ln 2$ to obtain

$(f \circ g)'(x) = ( f'\circ g) (x) \times g'(x) = \ln 2 \times e^{ x \ln 2 } = \ln 2 \times 2^x.$

Hence, the derivative of $2^x$ is $\ln 2 \times 2^x$. $_\square$

Find the derivative of $f(x) = e^x (x^2 + 1).$

Let $u = e^x$ and $v = x^2 + 1$ so that $f(x) = uv$. Then $u' = e^x$ and $v' = 2x.$

As $f(x) = uv,$ we get that $f'(x) = uv' + u'v,$ which implies

$f'(x) = e^x(2x) + (x^2 + 1)e^x = e^x(x^2 + 1 + 2x) = e^x(x + 1)^2.\ _\square$

Find the derivative of $f(x) = xe^x\sin x.$

Let $u(x) = x, v(x) = e^x,$ and $w(x) = \sin x$ so that $f(x) = u(x)v(x)w(x)$. Then $u'(x) = 1, v'(x) = e^x, w'(x) = \cos x.$

Now,

$\begin{aligned} f(x) & = u(x) \cdot v(x) \cdot w(x) \\\\ \Rightarrow f'(x) & = u'(x)v(x) w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x) \\ & = xe^x\cos x + xe^x\sin x + e^x\sin x \\ & = e^x(x\cos x + x\sin x + \sin x) .\ _\square \end{aligned}$

Find the derivative of $f(x) = 5^x + x^3e^x.$

Write $u(x) = 5^x$ and $v(x) = x^3e^x$ so that $f(x) = u(x) + v(x)$. Then $u'(x) = 5^x \cdot \ln 5$ and $v'(x) = \big(3x^2\big)e^x + x^3(e^x) = x^2e^x(3 + x)$.

Now,

$\begin{aligned} f(x) & = u(x) + v(x) \\\\ \Rightarrow f'(x) & = u'(x) + v'(x) \\ & = 5^x\cdot \ln 5 + x^2e^x(x + 3) .\ _\square \end{aligned}$

If $f(x) = 7^{x^3 + 3x}$ for $x > 0,$ then find $f'(x)$.

Write $u(x) = x^3 + 3x$ so that $f(x) = 7^{u(x)}$. Then $u'(x) = 3x^2 + 3$ .

Now,

$\begin{aligned} f(x) & = 7^{u(x)} \\\\ \Rightarrow f'(x) & = \big(7^{u(x)} \cdot \ln 7\big) \cdot u'(x) \\ & = 3(x^2 + 1)\big(7^{x^3 + 3x} \cdot \ln 7\big).\ _\square\end{aligned}$

**Cite as:**Derivatives of Exponential Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/derivatives-of-exponential-functions/