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Section Derivatives of the Exponential and Logarithmic Functions

We know how to take the derivative of \(f(x) = x^2,\) but what about \(g(x) = 2^x\) or \(h(x) = e^x?\) We wish to be able to differentiate exponential and logarithmic functions. If you need a review of these functions, then work through the problems in the appendix Exponential and Logarithmic Functions .

Problem 2.74.

Sketch graphs of \(f(x) = 2^x\) and \(g(x) = 3^x\) on the same coordinate axes labeling several points and being very careful to show which function is above the other at various points on the graph.

Example 2.75.

Defining \(e\) and finding the derivative of the function \(e^x\text{.}\)

Consider finding the derivative of \(f(x) = 2^x.\) Using limits to compute the derivative, we have:

\begin{align*} f'(x) \amp = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \;\;\; \mbox{by definition of derivative}\\ \amp = \lim_{h \rightarrow 0} \frac{2^{x+h}-2^x}{h} \;\;\; \mbox{by definition of} \; \; f\\ \amp = \lim_{h \rightarrow 0} \frac{2^x(2^h-1)}{h} \;\;\; \mbox{by algebra}\\ \amp = \lim_{h \rightarrow 0} 2^x\frac{2^h-1}{h} \;\;\; \mbox{by algebra}\\ \amp = 2^x \lim_{h \rightarrow 0} \frac{2^h-1}{h} \;\;\; \mbox{by the }\\ \amp = 2^x \; f'(0) \end{align*}

Consider finding the derivative of \(f(x) = 2^x.\) Using limits to compute the derivative, we have:

\begin{align*} f'(x) \amp = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \;\;\; \mbox{by definition of derivative}\\ \amp = \lim_{h \rightarrow 0} \frac{2^{x+h}-2^x}{h} \;\;\; \mbox{by definition of} \; \; f\\ \amp = \lim_{h \rightarrow 0} \frac{2^x(2^h-1)}{h} \;\;\; \mbox{by algebra}\\ \amp = \lim_{h \rightarrow 0} 2^x\frac{2^h-1}{h} \;\;\; \mbox{by algebra}\\ \amp = 2^x \lim_{h \rightarrow 0} \frac{2^h-1}{h} \;\;\; \mbox{by the }\\ \amp = 2^x \; f'(0) \end{align*}

We would now know the derivative of the function \(2^x\) if only we could determine \(f'(0) = \dsp{\lim_{h \rightarrow 0} \frac{2^{h}-1}{h}}.\) We can approximate \(f'(0) = \dsp{\lim_{h \rightarrow 0} \frac{2^{h}-1}{h}} \approx .69\) by using a limit table. Similarly, if we went through the same process for \(g(x) = 3^x\) we would see that \(g'(0) = \dsp{\lim_{h \rightarrow 0} \frac{3^{h}-1}{h}} \approx 1.09\) by using a limit table. Since the slope of \(f\) at 0 is less than 1 and the slope of \(g\) at 0 is greater than one, it makes sense that there should be an exponential function which has slope exactly 1 at 0. We define a number, called \(e\text{,}\) to be the number (between 2 and 3) so that \(\dsp{\lim_{h \rightarrow 0} \frac{e^{h}-1}{h}} = 1.\) Now, if we let \(j(x) = e^x\) then we have

\begin{align*} j'(x) \amp = \lim_{h \rightarrow 0} \frac{j(x+h)-j(x)}{h} \;\;\; \mbox{by definition of derivative}\\ \amp = \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} \;\;\; \mbox{by definition of} \; \; f\\ \amp = \lim_{h \rightarrow 0} \frac{e^x(e^h-1)}{h} \;\;\; \mbox{by algebra}\\ \amp = \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} \;\;\; \mbox{by definition of} \; \; f\\ \amp = \lim_{h \rightarrow 0} \frac{e^x(e^h-1)}{h} \;\;\; \mbox{by algebra}\\ \amp = \lim_{h \rightarrow 0} e^x\frac{e^h-1}{h} \;\;\; \mbox{by algebra}\\ \amp = e^x \lim_{h \rightarrow 0} \frac{e^h-1}{h} \;\;\; \mbox{by the }\\ \amp = e^x f'(0)\\ \amp =e^x. \end{align*}

This work results in a definition and a theorem.

Definition 2.76.

Let \(e\) be the number that satisfies \(\dsp \lim_{h \to 0} \frac{e^h-1}{h} = 1\text{.}\)

Definition 2.78.

Let \(g(x)=\ln(x)\) be the function that is the inverse of the function \(f(x) = e^x\text{.}\)

Algebra Reminders

  1. \(f(x)=e^x\) has domain all \(x \in \re\) and range all \(x > 0\)

  2. \(g(x)=\ln(x)\) has domain all \(x > 0\) and range all \(x \in \re\)

  3. \(e^{x+y} = e^x e^y\) and \({(e^x)}^{^y} = e^{xy}\) for all \(x,y \in \re\)

  4. \(\ln(xy) = \ln(x) + \ln(y)\) for all \(x>0\text{,}\) \(y>0\)

  5. \(\ln(x^y) = y\ln(x)\) for all \(y \in \re\) and \(x>0\)

  6. \(e^{\ln(x)} = x\) for all \(x > 0\)

  7. \(\ln(e^x)=x\) for all \(x \in \re\)

  8. \(\log_b(x) = \ln(x)/ln(b)\) for all \(b \in \re\) (except \(b\)=1) and for all \(x>0\)

Problem 2.79.

Compute and simplify the derivatives of the following functions:

  1. \(g(x) = e^x + x^2 + 2x\)

  2. \(f(x) = xe^x\)

  3. \(h(x) = x^3/e^x\)

  4. \(f(x) = e^{x^2}\)

  5. \(f(x) = 3x^2e^{x^4}\)

Problem 2.80.

Prove that if the graph of \(h(x) = f(x) - g(x)\) has a horizontal tangent line at \((c,h(c))\text{,}\) then the tangent line to f at \((c,f(c))\) is parallel to the tangent line to g at \((c,g(c))\text{.}\)

Problem 2.81.

Let \(g(x) = \ln(x)\text{.}\) Here are two clever ways to compute the derivative of \(g\text{;}\) do which ever you like better!

  1. Since we know that \(g(e^x)=x,\) differentiate both sides of this equation using the Chain Rule and then solve for \(g'(x).\)

  2. Apply the exponential function to both sides of the equations \(g(x)=\ln(x)\text{,}\) differentiate, and then solve for \(g'(x).\)

Problem 2.82.

Compute and simplify the derivative of the following functions:

  1. \(g(x) = \ln(7x-8)\)

  2. \(f(x) = x^2\ln(x)\)

  3. \(h(x) = x^3/\ln(3x^2 - x^{1/3})\)

Problem 2.83.

Let \(h(x) = 2^x\) and compute \(h'\) first by taking the natural log of both sides and then taking the derivative of both sides.

Problem 2.84.

Compute and simplify the derivative of the following functions:

  1. \(g(x) = 2^{7x-8}\)

  2. \(f(x) = 2^x - x^2\)

  3. \(h(x) = 3^{x^2}\)

Problem 2.85.

Use the change-of-base formula, \(\log_b(x) = \ln(x)/\ln(b)\text{,}\) to compute the derivative of \(L(x) = \log_3(x^5-2x).\)

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