If $f(x)$ is a continuous and differentiable function which satisfies the function equation If $$2f (x) = f(xy) + f\left(\frac xy\right)\quad \forall x,y \in \mathbb{R}^{+}$$ and $f'(1)=1$ then find $f(x)$
I can see that $f(x)=\ln(x)$ is one function which satisfies all the properties but how can it be proved? I tried using first principle of differentiation but wasn't able to obtain the function. Could someone help me with this?
Let's differentiate your equation with respect to $y$
$$0=xf^{\prime}\left(xy\right)-\dfrac{x}{y^{2}}f^{\prime}\left(\dfrac{x}{y}\right) \tag{1}$$
This is allowed since $f$ is differentiable. Now set $x=y$ and use $f^{\prime}(1)=1$ to have
$$0=xf^{\prime}\left(x^{2}\right)-\dfrac{1}{x} \tag{2}$$
or if we define $u=x^{2}$
$$f^{\prime}(u)=\dfrac{1}{u} \tag{3}$$
Thus $f\left(u\right)=\ln(u)+C$ for $C\in\mathbb{R}$.
Edit 1: Following the answer by @Bumblebee, we can solve this equation in general by not specifying $C_{1}=f^{\prime}(1)$. In this case, Eq. $(3)$ is in fact
$$f^{\prime}(u)=\dfrac{C_{1}}{u} \tag{4}$$
with a solution
$$f(u)=C_{1}\ln(u)+C_{2} \tag{5}$$