# 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:

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

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,

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

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,

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

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,

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

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

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