The functional equation $f(xy)=f(x)+f(y)+\frac{x+y-1}{xy}$ with a hint of differentiability Okay so I got this question in a rehearsal test today, and got completely stumped.
Let $f$ be a differentiable function satisfying the functional rule $f(xy)=f(x)+f(y)+\frac{x+y-1}{xy}$, $\forall$ $x,y>0$ and $f'(1)=2$.The question asked some more questions regarding $f(x)$,but I am unable to find $f(x)$.
My try:I tried to treat $y$ as a constant and then differentiated both sides of the equation,making use of the fact that $f'(1)=2$ ; but that got me nowhere.Someone kindly help.
Thanks in advance.
 A: Your try gives
$$yf'(xy) = f'(x) -\frac{1}{x^2} + \frac{1}{x^2y},\tag{1}$$
for all $x,y > 0$. Choosing $x = 1$ in $(1)$, we obtain
$$yf'(y) = 2 - 1 + \frac{1}{y},$$
or
$$f'(y) = \frac{1}{y} + \frac{1}{y^2}.\tag{2}$$
From that, we can easily determine $f$ up to addition of a constant, and we can determine the constant from the rule e.g. by setting $x = y = 1$ there.
An alternative way to find the function is to rewrite the functional equation as
$$f(xy) + \frac{1}{xy} = \biggl(f(x) + \frac{1}{x}\biggr) + \biggl(f(y) + \frac{1}{y}\biggr),$$
and if one knows the continuous functions satisfying the functional equation $g(xy) = g(x) + g(y)$, $f$ is easily determined.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\mrm{f}}$ is differentiable.
  $\ds{\mrm{f}\pars{xy} = \,\mrm{f}\pars{x} + \,\mrm{f}\pars{y} + {1 \over y} +
{1 \over x} - {1 \over xy}\,,\quad
x, y \in \mathbb{R}_{\ >\ 0}\quad\mbox{and}\quad\mrm{f}'\pars{1} = 2}$.

$$
\bbx{\quad\mbox{Note that}\quad\mrm{f}\pars{1} = -1\quad
\pars{~\mbox{from the 'original functional equation'}~}\quad}
$$


*

*Derive both members, of the above functional equation, respect of $\ds{x}$:
\begin{equation}
\mrm{f}'\pars{xy}y =
\,\mrm{f}'\pars{x} - {1 \over x^{2}} + {1 \over x^{2}y}
\implies
\mrm{f}'\pars{xy}xy =
\,\mrm{f}'\pars{x}x - {1 \over x} + {1 \over xy}
\label{1}\tag{1}
\end{equation}

*Derive both members, of the above functional equation, respect of $\ds{y}$:
\begin{equation}
\mrm{f}'\pars{xy}x =
\,\mrm{f}'\pars{y} - {1 \over y^{2}} + {1 \over xy^{2}}
\implies
\mrm{f}'\pars{xy}xy =
\,\mrm{f}'\pars{y}y - {1 \over y} + {1 \over xy}
\label{2}\tag{2}
\end{equation}
\eqref{1} and \eqref{2} lead to:
\begin{align}
&\,\mrm{f}'\pars{x}x - {1 \over x} = \,\mrm{f}'\pars{y}y - {1 \over y} = \alpha\,,
\qquad\pars{~\alpha:\ \mbox{a}\ x\ \mbox{and}\ y\ \mbox{independent constant}~}
\end{align}


$\ds{\alpha = \,\mrm{f}'\pars{1} \times 1 - {1 \over 1} = 1}$.

Now, you are left with
$$
\,\mrm{f}'\pars{x} = {1 \over x} + {1 \over x^{2}}\,,\quad\mrm{f}\pars{1} = -1
\implies \bbox[#ffe,10px,border:1px dotted navy]{\ds{\mrm{f}\pars{x} = \ln\pars{x} - {1 \over x}}}
$$
