Is this proof that the derivative of $\ln(x)$ is $1/x$ correct?

\begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align}

\begin{align} &= \lim_{h\to0}{\ln(1 + \frac{h}{x})}^\frac{1}{h} \\ \end{align} Now this is the part I'm asking about, because I see most people set the reciprocal of my limit, but this one seems to work: \begin{align} n=\frac{x}{h}, h=\frac{x}{n}, \frac{1}{h} = \frac{n}{1}*\frac{1}{x} \end{align} So when h tends towards 0, n tends towards infinity. \begin{align} &= \lim_{n\to\infty}{\ln\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\ \end{align} Here we have \begin{align} {\ln\lim_{n\to\infty}\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\ \end{align} \begin{align} =\frac1x\ln(e) = \frac1x \end{align} This is my first post, I was just curious. Please don't berate me if it's a silly question.

Solution 1:

I think this is OK. You should acknowledge that at one point you are using the fact that the function raising something to a fixed power is continuous: $$ \lim (\text{something}^a)= (\lim (\text{something}) )^a . $$

Moreover, the usual definition of exponentiation for an arbitrary exponent $a$ depends on the natural logarithm: $$ \text{something}^a = \exp(a (\ln \text{something})) $$ so there's a lot of serious analysis (sometimes skipped over the first time through in calculus) in the correct algebraic manipulations you're doing.

That said, this is a strange exercise. Presumably you have defined $\ln$ as the inverse of exponentiation, so that $$ \exp(\ln(x)) = x . $$ Then the formula for the derivative of $\ln$ follows from the chain rule.

Solution 2:

yeah your answer looks correct , I'd probably do it this way tho ;

$\lim_{h\to0}\frac1h\ln\big(1+\frac hx) \\= \lim_{h\to0}\frac xh \frac1x\ln\big(1+\frac hx) \\= \frac1x\lim_{h\to0}\ln\big(1+\frac 1{\frac xh})^\frac xh\\ = \frac1x\ln(e) \\= \frac1x$