Implicit differentiation to find derivatives of a function whose only defined by its derivative 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.
 A: 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!
