According to Wikipedia, one common definition of the natural logarithm is that:
$$
\ln (x) = \int_{1}^{x} \frac{1}{t} dt
$$
The article then goes on to say that because of the first FTC, we can deduce that:
$$
\frac{d}{dx}\left(\ln (x)\right)= \frac{1}{x}
$$
This doesn't make sense to me. Although I agree that the derivative is indeed equal to $\frac{1}{x}$, I don't understand how that follows from the first FTC.
To my knowledge, the first FTC tells us that definite integrals can be computed using indefinite integrals. If we have $f(x)$, and $F(x)=\int f(x) dx$, then
$$
\int_{a}^{b} f(x) = F(b)-F(a)
$$
I understand that this is a significant result because definite integrals are defined as equalling as the area under the graph between $a$ and $b$, not with some formula that involves indefinite integrals.
If this is the case, then applying the first FTC to the problem at hand seems to only get us so far:
$$
\ln (x) = \int_{1}^{x} \frac{1}{t} dt
$$
And then we are stuck, because though we know that $f(t)= \frac{1}{t}$, we haven't shown that $F(x) = \ln(x)$. The only thing that we have shown is that $\ln (x) = \int_{1}^{x} \frac{1}{t} dt$. What am I missing?
Best Answer
I guess that you are missing that
$$F(x)=\ln(x)$$ by definition, and $$f(x)=\frac1x.$$
So by the FTC,
$$\int_1^x\dfrac{dt}t=F(x)-F(1)=\ln(x)$$
and as $F$ is an antiderivative of $f$,
$$\frac{d\ln x}{dx}=F'(x)=f(x)=\frac1x.$$