Show that the derivative of ln(x) is 1/x.

[Greg]

Show that the derivative of $\ln(x)$ is $1/x$.

 

[MarkFL]

My solution:

Spoiler

We begin by writing:

$$y=\ln(x)$$

Convert from logarithmic to exponential form:

$$x=e^y$$

Differentiating w.r.t $y$, we obtain:

$$\d{x}{y}=e^y=x\implies \d{y}{x}=\frac{1}{x}$$

 

[Dethrone]

A proof that I'm not sure if rigorous -- I just hacked my way through:

Spoiler

\begin{equation}
\begin{split}
\ln'(x)&=\lim_{h \to 0}\frac{\ln(x+h)-\ln(x)}{h}\\
&=\lim_{h \to 0} \frac{\ln \left(1+\frac{h}{x}\right)}{h}\\
&= \lim_{h \to 0} \frac{\ln'(1+\frac{h}{x})\cdot \frac{1}{x}}{1} \text{ (l'hopital)}\\
&= \frac{\ln'(1)}{x}
\end{split}
\end{equation}

To find $\ln'(1)$, we can substitute $x=1$ into the definition of the derivative, which yields $1/h \cdot [\ln(1+h)-\ln(1)]$ as $h \to 0$, applying l'hopital on $h$ gives us $1/(1+h) \to 1$ as $h \to 0$.


Oops, the derivative of $\ln x$ is so second nature that I applied it without realizing (Rofl). Let's try again:
\begin{equation}
\begin{split}
\ln'(1)&=\lim_{h \to 0}\frac{1+h}{h}\\
&= \lim_{h \to 0} \ln(1+h)^{1/h}\\
&=\ln e\\
&=1
\end{split}
\end{equation}

Therefore,
$$\ln'(x)=\frac{1}{x}$$

 

[I like Serena]

My take on it. (Smile)

Spoiler

There are a number of equivalent definitions for $\ln$.

According to wiki, which is also how I originally learned it, the definition of $\ln$ is given as:
$$\ln(x)\overset{\small\text{def}}=\int_1^x \frac 1x\,dx$$
Then, by the first fundamental theorem of calculus:
$$\ln'(x)=\frac 1x$$

 

[kaliprasad]

My solution

Spoiler

$\ln'(x)=\lim_{h \to 0}\frac{ln(x+h) - ln (x)}{h}$
$= \lim_{h \to 0}\frac{ln\frac{x+h}{x}}{h}$
$= \lim_{h \to 0} \ln(1+\frac{h}{x})^{\frac{1}{h}}$
$= \lim_{h \to 0} \ln(1+\frac{h}{x})^{\frac{\frac{x}{h}}{x}}$
$= \lim_{h \to 0} (\ln(1+\frac{h}{x})^{\frac{x}{h}})^{\frac{1}{x}}$
$= ln ( \lim_{h \to 0} (1+\frac{h}{x})^{\frac{x}{h}})^{\frac{1}{x}}$
$= ln (e^{\frac{1}{x}})$ from above and definition of $e$
$= \frac{1}{x}$

 

[HallsofIvy]

Does Wiki really have $$\int_1^x \frac{1}{x}dx$$?

That's just bad notation! It should be $$\int_1^x \frac{1}{t}dt$$.
Don't use the same symbol, "x", for two different things.

 

[I like Serena]

Does Wiki really have $$\int_1^x \frac{1}{x}dx$$?

That's just bad notation! It should be $$\int_1^x \frac{1}{t}dt$$.
Don't use the same symbol, "x", for two different things.

Actually, it doesn't. I just thought it looked prettier in the context of this problem.

 

[Greg]

Spoiler

$$\int\frac1x\,dx,\quad e^u=x,e^u\,du=dx,\,u=\ln(x)$$

$$\int\,du=u+C$$

$$\int\frac1x\,dx=\ln(x)+C$$