how do you differentiate $\ln(x)$ using the difference quotient.

For finite non-zero $x,$ $$\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}h=\dfrac1x\cdot\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}{\dfrac hx}=?$$

In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\ge \log(x)\le x-1\tag1$$

Hence, using $\log(x+h)-\log(x)=\log\left(1+\frac hx\right)$, we from $(1)$ we have

$$\frac{\frac1x}{1+\frac hx}\le \frac{\log\left(1+\frac hx\right)}{h}\le \frac1x\tag2$$

Applying the squeeze theorem to $(2)$ we find the coveted limit

$$\lim_{h\to 0}\frac{\log(x+h)-\log(x)}{h}=\frac1x$$

If the definition of $\ln x$ is

$$\ln x = \int_1^x \frac{1}{t} \; dt$$

Then you can write your limit as

$$\lim_{h \to 0 } \frac{1}{h} \int_x^{x+h} \frac{1}{t} \; dt.$$

Then note that for very small $h$, the area represented by the integral is nearly a rectangle of width $h$ and height $1/x$, so the limit equals

$$\lim_{h \to 0} \frac{1}{h}(h/x) = \frac{1}{x}.$$