How does the Fundamental Theorem of Calculus (FTC) tell us that $\frac{d}{dx}\left(\ln (x)\right)= \frac{1}{x}$?

Question

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?

Answer 1

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 $f$,

$$\frac{d\ln x}{dx}=F'(x)=f(x)=\frac1x.$$

Answer 2

There are two parts to the fundamental theorem of calculus: the other part is that, for $f$ continuous on $[a,b]$: $$\frac{d}{dx}\int_a^x f(t)\,dt=f(x).$$ So now it should be clear that, by definition: $$\frac{d}{dx}\int_1^x\frac{1}{t}\,dt=\frac{1}{x}.$$

Answer 3

$ P(x)\ =\ _{a(x)}\int^{b(x)}\ f(t)dt $

Let $F(x)\ =\ \int f(x)$

Hence $ P(x)\ =\ F(b(x))\ -\ F(a(x))$

$\frac{d(P(x))}{dx}=P'(x)\ =\ F'(b(x)).b'(x)\ +\ F'(a(x)).a'(x)$

$P'(x) = f(b(x)).b'(x)+f(a(x)).a'(x)$

And in your question $a(x)=1$ , $b(x)=x$ and $f(x)=\frac{1}{x}$

Answer 4

If you define a function as an integral, it's easy to see that its derivative will be the integrand. However, I think this is a rather bad way of defining the natural logarithm and its derivative, so I'll illustrate what is in my opinion the best way to approach this problem.

First, we'll start with the canonical definition of $e$, $\lim_{n \to \infty}{(1+\frac{1}{n})^n}=e$. Well, before we can do this we must show that the limit exists. This is fairly easily shown using the convergence of monotone bounded sequences. Next, using the limit definition, we can show that $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$. This is a challenging, but very rewarding and beautiful proof that is sadly never shown in almost all first year calculus courses. With this in mind, we'll define the natural logarithm $\ln:x \mapsto \ln(x)$ as the inverse of the exponential function, that is it is the unique function with the property that $\ln(e^x)=e^{\ln(x)}=x.$ Now we'll use the following property of inverse functions: Supposing $f$ is a differentiable function with an inverse $f^{-1}$, then

$$\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$$ This can be shown easily with the chain rule. Substitute $f(x)=e^x$ and $f^{-1}(x)=\ln(x)$ and using the property that $f'(x)=f(x)$, we have $$\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{e^{\ln(x)}}=\frac{1}{x}.$$

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Added for fun since this proof is gorgeous and I like typesetting math: A proof that $\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x$. Let's begin.

$$e^x=\lim_{n \to \infty}{\left(1+\frac{1}{n}\right)^{nx}}=\lim_{n \to \infty}{\left(1+\frac{x}{n}\right)^{n}}$$ This looks like a perfect time to use the binomial theorem. So then, $$e^x=\lim_{n \to \infty}{\sum_{k=0}^n{{}_n \mathrm{C}_k\cdot1^{n-k}\cdot\left(\frac{x}{n}\right)^{k}}}=\lim_{n\to\infty}{\sum_{k=0}^n{\frac{n!}{(n-k)!k!}\frac{x^k}{n^k}}}$$ $$=\lim_{n\to\infty}{\left(1+x+\frac{n(n-1)}{n^2}\frac{x^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{x^3}{3!}+...\right)}=\sum_{k=0}^{\infty}{\frac{x^k}{k!}}.$$ So then, $$\frac{\mathrm{d}}{\mathrm{d}x}e^x=\lim_{h\to 0}{\left(\frac{\sum_{k=0}^{\infty}{\frac{(x+h)^k}{k!}-\sum_{k=0}^{\infty}{\frac{x^k}{k!}}}}{h}\right)}=\frac{1}{k!}\lim_{h\to 0}{\left(\frac{1}{h}\sum_{k=0}^{\infty}{\left((x+h)^k-x^k\right)}\right)}$$ Let's examine the thing inside the sum. $$(x+h)^k-x^k=(x^k+khx^{k-1}+{}_k \mathrm{C}_2h^2x^{k-2}+...+h^k)-x^k$$ $$=h(kx^{k-1}+{}_k \mathrm{C}_2hx^{k-2}+...+h^{k-1})$$

Therefore, $$\frac{\mathrm{d}}{\mathrm{d}x}e^x=\frac{1}{k!}\lim_{h\to 0}{\left(\sum_{k=0}^{\infty}{\left(kx^{k-1}+{}_k \mathrm{C}_2hx^{k-2}+...+h^{k-1}\right)}\right)}$$ $$=\sum_{k=0}^{\infty}{\frac{kx^{k-1}}{k!}}=\sum_{k=0}^{\infty}{\frac{x^k}{k!}}=e^x.$$