Derivative of logarithmic functions

Theorem 1

Consider the logarithmic function $f(x)=\log_a(x)$ where $a$ is the base (e.g. $f(x)=\ln(x)$ , or $f(x)=\log_2(x)$ ). The following is true:

Equation 1

f(x)=\log_a(x) \overset{\prime}{\rightarrow} f'(x)=\frac{1}{\ln(a)\cdot x}

Derivative of Logarithmic Function

In particular it follows

Equation 2

f(x)=\ln(x) \overset{\prime}{\rightarrow} f'(x)=\frac{1}{x}

Proof

We just give a plausibility argument. Sketching the derivative of $f(x)=\ln(x)$ , we see that its graph indeed looks similar to the graph of the function $\frac{1}{x}$ .

So we can argue (but it is not a proof, of course), that the derative of $\ln(x)$ is $f^\prime{x}=\frac{1}{x}$ . And this turns out to be true. The more general case for $f(x)=\log_a(x)$ will be proven later with the chain rule.

Example 1

$f(x)=\log_2(x) \rightarrow f'(x)=\frac{1}{\ln(2)\cdot x}=\frac{1}{0.693...\cdot x}$
$f(x)=\ln(x) \rightarrow f'(x)=\frac{1}{\ln(e)\cdot x}=\frac{1}{x}$

A more formal proof of the theorem is given in the exercise (uncollapse) but can be skipped. The graphical argument is more important.

Show

Prove that the derivative of $f(x)=\log_2(x)$ is $f'(x)=\frac{1}{\ln(2)\cdot x}$ . To do so, answer the questions below. You can also use the rule that the derivative of $g(x)=2^x$ is $g'(x)=\ln(2)\cdot 2^x$ . Let's use some small exercises for this:

Exercise 1

Consider a point $P(x\vert y)$ . If $P$ is reflected about the diagonal, what are the coordinates of the new point $P'$ ?

Solution

Reflect some points, and you will see that the order of the coordinates switches: $P(x\vert y) \rightarrow P'(y\vert x)$

Exercise 2

A straight line $t$ with slope $a$ . If $t$ is reflected about the diagonal, what is the slope of the new line $t'$ ?

Solution

The slope of $t'$ is $a'=\frac{1}{a}$ .

To see this, pick to points $A(x_A\vert y_A)$ and $B(x_B\vert y_B)$ on $t$ and form the slope triangle between these two points. The slope of $t$ is $a=\frac{y_B-y_A}{x_B-x_A}$ . The reflected points $A'$ and $B'$ are on $t'$ and have the coordinates switched, that is, $A'(y_A\vert x_A)$ and $B'(y_B \vert x_B)$ . Thus, the slope of $t'$ is $a'=\frac{x_B-x_A}{y_B-y_A}=\frac{1}{\frac{y_B-y_A}{x_B-x_A}}=\frac{1}{a}$ .

Exercise 3

The graphs of the functions $f(x)=\log_2(x)$ and $g(x)=2^x$ are shown below. Observe that if we reflect the graph of $f$ about the diagonal, we obtain the graph of $g$ . Can you show this with a calculation? For example, take the point $A$ with coordinate $x=2.5$ on the graph of $f$ . Why must the reflected point be on the graph of $g$ , and what are its exact coordinates?

Solution

Note that a point $P(x\vert y)$ is on the graph of a function $l$ , if $y=l(x)$ . As $A$ is on the graph of $f$ , it must have the coordinates $A(2.5\vert f(2.5))=A(2.5\vert \log_2(2.5))$ .

The reflected point has therefore the switched coordinates $A'(\log_2(2.5)\vert 2.5)$ . To check if this point is on $g$ , we have to show that $g(\log_2(2.5))=2^{\log_2(2.5)}=2.5$ . And this is indeed the case, because if we set $\log_2(2.5)=u$ it follows from the definition of the logarithm that $2^{u}=2.5$ . So we have $2^{\log_2(2.5)}=2^{u}=2.5$

Exercise 4

Put all the pieces together. Can you show that $f'(2.5)=\frac{1}{\ln(2) \cdot 2.5}$ ? More generally, can you show that $f'(x)=\frac{1}{\ln(2)\cdot x}$ ? Hint: Use the already proven rule $g'(x)=\ln(2)\cdot 2^x$

Solution

Consider the tangent $t$ to the graph of $f$ at $A(2.5\vert \log_2(2.5))$ and denote its slope by $a$ , that is, $a=f'(2.5)$ . We want to find the value of $a$ .

To do so note that the reflected tangent $t'$ passes through the point $A'(\log_2(2.5)\vert 2.5)$ (see Q1 and Q3) and is the tangent $t'$ to the graph of $g$ . As the slope of $t'$ is $a'=g'(\log_2(2.5))=\ln(2)\cdot 2^{\log_2(2.5)}=\ln(2)\cdot 2.5$ (see hint), and because $a=\frac{1}{a'}$ (see Q2) we get $a=\frac{1}{\ln(2)\cdot 2.5}$ .

The argumentation for the general rule is similar.

Exercise 5

Consider the function $f(x)=\log_3(x)-x^2+3$ .
1. Sketch the graph of $f$ (you can use the calculator).
2. Find the coordinates of the highest point $P$ of the graph.
A straight line $l$ passes through the origin and touches the graph of the function $f(x)=\log_5(x)$ at point $P$ .
1. Find the coordinates of $P$ .
2. Find the angle of intersection between $l$ and the $x$ -axis.

Solution

1. To draw the graph, use a table of values (hint: use powers of $3$ for the $x$ -values). The graph is shown below.
2. The tangent at the highest point of graph $f$ has slope $0$ , that is, we have to find $x$ with $f'(x)=\frac{1}{\ln(3)x}-2x=0$ . It follows $x=\sqrt{\frac{1}{2\ln(3)}}=0.675$ and thus $\underline{P(0.675\vert 2.187)}$ .
Assume $P$ has the (unknown) $x$ -coordinates $u$ , thus we have $P(u\vert\log_5(u))$ . We need to find $u$ . Create a figure. You will note that we can express the slope of the line $l$ in two different ways:
- Because $l$ passes through the origin and through $P(\vert\log_5(u))$ , we can form a slope triangle with $\Delta x = u$ and $\Delta y=\log_5(u)$ . Thus $l$ has slope $a=\frac{\Delta y}{\Delta x}=\frac{\log_5(u)}{u}$
- Because $l$ is the tangent to the graph of $f$ at $x=u$ , we also have $a=f'(u)=\frac{1}{\ln(5)u}$ .
Taken together, we have two expression for $a$ and therefore an equation for $u$ :
$\frac{\log_5(u)}{u} = \frac{1}{\ln(5)u}$
Multiplying both sides by $u$ , we get
$\log_5(u) = \frac{1}{\ln(5)}$
In other words, $5^{1/\ln(5)}=u$ , thus $u=2.718$ and $\underline{P(2.718\vert 0.621)}$ . The slope of $l$ is $a=\frac{1}{\ln(5)\cdot 2.718}=0.229$ , the angle of intersection between $l$ and the $x$ -axis is therefore $\arctan(0.229)=\underline{12.877}$