8.4: Derivatives of Exponential Functions

Now that we can take limits with exponential functions, the next step is to find its derivative:

If \(f(x)=b^x\), and \(b>0\) then:

\( f'(a)=\displaystyle{ \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=\lim_{h\to 0} \frac{b^{a+h}-b^a}{h}=\lim_{h\to 0} \frac{b^{a}\cdot b^h-b^a}{h}=b^a\cdot \textcolor{red}{\lim_{h\to 0} \frac{b^h-1}{h} } }\)

The remaining limit is not easy to solve, but lets name it something: \(\textcolor{red}{L(b)=\displaystyle{\lim_{h\to 0} \frac{b^h-1}{h}}}\)

There is at least one value of \(b\) that we can easily plug into \(L\):

\(L(1)=\displaystyle{ \lim_{h\to 0} \frac{1^h-1}{h}=\lim_{h\to 0} \frac{1-1}{h}=\lim_{h\to 0} 0=0}\)
( Thats not too surprising, considering \(f(x)=1^x=1\) is a constant function, so \(f'(x)=0\) )

Sadly, there are no other easy values of \(b\) to plug in, but \(L(b)\) is still a continuous function.

Lets plug in a few nice numbers and see what comes out:

Clearly \(L(2)<1\) and \(L(3)>1\), therefore by the Intermediate Value Theorem, we know there is some value \(c\in(2,3)\) such that \(L(c)=1\)

Since \(\displaystyle{\lim_{h\to 0} \frac{c^h-1}{h}=1}\), we can estimate it by picking \(h=\frac{1}{n}\) to get:

\(\displaystyle{ \frac{c^{\frac{1}{n}}-1}{\frac{1}{n}}\approx 1 \longleftrightarrow c\approx \left(1+\frac{1}{n}\right)^n }\)

This value of \(c\) is referred to as Euler's Number, written as \(e\):

\(e= \displaystyle{\lim_{n\to \infty} \left( 1+\frac{1}{n}\right)^n\approx 2.7182}\)

Another property of \(L(b)\) can help us when \(b=e^r\) for some constant \(r\ne 0\):

\(\displaystyle{ L(e^r)=\lim_{h\to 0} \frac{(e^r)^h-1}{h}=\lim_{h\to 0} \frac{e^{rh}-1}{h}\cdot \textcolor{red}{\frac{r}{r}}=r\cdot \lim_{\textcolor{purple}{rh}\to 0} \frac{e^{\textcolor{purple}{rh}}-1}{\textcolor{purple}{rh}}=r\cdot L(e)=r}\)

The end result then is for \(f(x)=(e^r)^x=e^{rx}\) we have \(f'(a)=(e^r)^aL(e^r)=re^{ar}\)

We will determine how to deal with bases that aren't of the form \(e^r\) in a later chapter.

For now we can combine these functions with others for more derivative practice:

Example \(\PageIndex{1}\)

Find the derivative of each function below involving exponentials.

(a) \(q(t)=e^{2t}\tan(t)\)

(b) \(u(s)=\displaystyle{\frac{e^s-1}{e^{-s}+s}}\)

Solution

(a) \(q'(t)=\left(e^{2t}\right)'(\tan(t))+\left(e^{2t}\right) (\tan(t))' =\left(2e^{2t}\right)(\tan(t))+\left(e^{2t}\right) (\sec^2(t))e^{2t} \left(2\tan(t) + \sec^2(t)\right)\)

(b) \(\displaystyle{u'(s)=\frac{(e^s-1)'(e^{-s}+s)-(e^s-1)(e^{-s}+s)'}{(e^{-s}+s)^2}=\frac{(e^s)(e^{-s}+s)-(e^s-1)(-e^{-s}+1)}{(e^{-s}+s)^2}=\frac{3+(s-1)e^s-e^{-s} }{(e^{-s}+s)^2}} \)

These exponential functions can sometimes end up making easily recognizable patterns when taking multiple derivatives (which is something that will be useful in the distant future):

Example \(\PageIndex{2}\)

Find the formula for \(f^{(n)}(x)\) when \(f(x)=xe^x\)

Solution

We start by taking the first few derivatives:

\(f'(x)=(x)'(e^x)+(x)(e^x)'=1\cdot e^x + x \cdot e^x = (x+1)e^x\)
\(f^{(2)}(x)=(x+1)'(e^x)+(x+1)(e^x)'=1\cdot e^x + (x+1) \cdot e^x = (x+2)e^x\)
\(f^{(3)}(x)=(x+2)'(e^x)+(x+2)(e^x)'=1\cdot e^x + (x+2) \cdot e^x = (x+3)e^x\)
\(f^{(4)}(x)=(x+3)'(e^x)+(x+3)(e^x)'=1\cdot e^x + (x+3) \cdot e^x = (x+4)e^x\)

It is clear to see the pattern is \(f^{(n)}(x)=(x+n)e^x\)

And we can even use these new exponential derivatives to make piecewise functions differentiable:

Example \(\PageIndex{3}\)

For what \(\alpha,\beta\) is \(g(x)=\left\{ \begin{array} {ll} e^{\alpha x}+\beta e^{x} & x> 0 \\ 5x-6 & x\le 0 \end{array} \right.\) differentiable at \(x=0\)?

Solution

First, we need \(g(x)\) to be continuous:

\(\displaystyle{ \lim_{x\to 0^+} g(x) =\lim_{x\to 0^+} e^{\alpha x}+\beta e^x = e^{\alpha \cdot 0}+\beta e^0=1+\beta}\)

\(\displaystyle{ \lim_{x\to 0^-} g(x) =\lim_{x\to 0^+} 5x-6 = 5\cdot 0-6=-6}\)

and since \(g(0)=-6\), we just need \(1+\beta=-6\)
Therefore if \(\beta=-7\), \(g(x)\) will be continuous at \(x=0\)

We can find \(g'(x)\) everywhere except at \(x=0\) by differentiating each piece:

\(g'(x)=\left\{ \begin{array} {ll} \alpha e^{\alpha x}+\beta e^{x} & x> 0 \\ 5 & x< 0 \\ ? & x=0 \end{array} \right.\)

To be differentiable at \(x=0\), the slope needs to match on both sides of \(0\):

\(\displaystyle{ \lim_{x\to 0^+} g'(x)=\lim_{x\to 0^+} \alpha e^{\alpha x}+\beta e^x=\alpha \cdot e^{\alpha\cdot 0}+\beta\cdot e^0}=\alpha+\beta\)

\(\displaystyle{ \lim_{x\to 0^-} g'(x) =\lim_{x\to 0^-} 5 = 5}\)

So to be differentiable we need \(\alpha+\beta=5\)

Since we already solved \(\beta=-7\), this means we need \(\alpha=12\)

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