I have the following exercise and I think I know how to solve it, but I need to base the development better.
Let $$\begin{array}{lrl}L:&(0,+\infty)&\longrightarrow&\mathbb{R}\\& x&\longmapsto&\displaystyle\int_{1}^{x}\dfrac{1}{t}dt\end{array}$$ a) Prove that $L$ is derivable throughout its domain and calculate its derivative
My solution: Here, I can say that $$\int_{1}^{x}\frac{1}{t}dt=\ln(x)$$ since the domain of the function is defined for $x>0$ then it is derivable for the same interval. And using the fundamental calculation theorem, I can get the respective derivative, right?
The doubt that I have is how I can formally prove it.
b) Prove that for all $x,y \in \textrm{dom (L)}$ it is verified that $$L(xy)=L(x)+L(y)$$ clearly it is verified, because it is a property of the logarithms; but, I want to know how to prove it.
I think this task should show how to derive the existence of the logarithmic function.
a) By the fundamental theorem of calculus $L$ is differentiable with $L'(x)=\frac1x$ for all $x>0$. b) Fix $y$ and put $f(x):=L(xy)-L(x)-L(y)$. Then $f'(x)=\frac{y}{xy}-\frac1x-0=0$ for all $x$. Moreover $f(1)=L(y)-L(1)-L(y)=0$. Thus $f(x)=0$ for all $x$. So the assertion in b) holds true for all $x$ with arbitrary $y$.