Showing that $M(xy) = M(x) + M(y)$ where $y$ is a constant for $M(x) = \int_1^x {1\over t}{d \over dt}$

Question

I'm given a function $M: (0, \infty) \to R$ such that $M(x) = \int_1^x {1\over t}{d \over dt}$. I'm told I have to use differentiation to solve this problem.

I differentiated $M(xy)$ first, for some constant y, getting ${d\over dx}M(xy) = {1 \over x}.$ So an antiderivative for $M(xy)$ is $M(x) + c$. If I let $c = M(y)$ i get that ${d\over dx}M(xy) = {d \over dx}(M(x) + M(y))$. However, just becaue two functions' derivatives are the same, does not mean that the functions are the same as well. So this is where I get stuck.

Answer 1

$\dfrac{d}{dx}M(xy)=yM'(xy)=y\cdot\dfrac{1}{xy}=\dfrac{1}{x}$, so $\dfrac{d}{dx}M(xy)=\dfrac{d}{dx}M(x)$, then $M(xy)=M(x)+C$, now $M(y)=M(1)+C=C$.

Note that $\dfrac{d}{dx}M(xy)$ should be interpreted as $\dfrac{d}{dx}M\circ\varphi(x)$, where $\varphi(x)=xy$.

Answer 2

To be fair, differentiation is not needed, only a change of variables: $$ M(xy)-M(x)=\int_x^{xy}\frac{dt}{t}=\int_1^y \frac{du}{u}=M(y), $$ where we let $u=t/x$.