A question I have asks if $K(x)$ satisfies $K(1)=0$ and $K'(x)={1\over{x}}$ then show:
If $f(x)=K(10x)$ then $f'(x)={1\over{x}}$
So Im not sure if I have to prove it or do something else but this is what I did.
$$
f'(x)=K(10x)^0\cdot K'(10x) \cdot 10
\\
f'(x)={1\over{10x}}\cdot 10={1\over{x}}
$$
The other questions are similar to this so if this is okay then im sure I did the rest okay as well. Then it asks:
$$
Let \ g(x)=K(x^2+1)\ \ \ \ find \ g'(x)
$$
So I do:
$$
g'(x)=K'(x^2+1)\cdot 2x = {2x\over{x^2+1}}
$$
At this point I havent used the $K(1)=0$ so im not sure if im doing this right. The last sub question asks:
If $a,b >0$ show $K(ab) = K(a)+K(b)$. Which I have no idea where to start.
Any feedback appreciated, thanks.
Your calculations seems to be totally fine in my opinion.
Note that instead of using implicit differentiation, it is possible to solve the differential equation
$$ K'(x) = \frac{1}{x},\qquad K(1)=0, $$ and use the explicit form of $K$ in your calculations. But your method is more time efficient and does the job perfectly fine. Although the last task is easier to solve with an explicit expression of $K$, so here we go;
$$ \frac{dK}{dx} = \frac{1}{x} \qquad \Rightarrow \qquad K(x) = \int \frac{1}{x} dx = \ln x + \alpha, \qquad \forall x\in\mathbb{R}^{+}\setminus \lbrace 0\rbrace, $$ where $\alpha \in \mathbb{R}$ is some constant. We know that $K(1) = 0$, i.e. $$ k(1) = \underbrace{\ln 1}_{=0} + \alpha \qquad \Rightarrow \qquad \alpha \equiv 0. $$ Thus, it must hold that $$ K(x) = \ln x,\qquad \forall x\in\mathbb{R}^{+}\setminus \lbrace 0\rbrace. $$
For this reason, the proof becomes pretty straightforward; $$ K(ab) = \ln ab = \ln a + \ln b = K(a) + K(b). $$
Sorry for the previous answer, sleep deprivation is a bad thing... =D
Good luck!