Cauchy mean value theorem, finding the function $\frac{f(c) - f(a) }{g(b) -g(c)} = \frac{f'(c)}{g'(c)}$. 
If $f'(x)$ and $g'(x)$ exist for all $x\in[a,b]$ and if $g'(x)$ does not vanish any where on $(a,b)$ then prove that for some $C\in(a,b)$ 
  $$\frac{f(c) - f(a) }{g(b) -g(c)}  = \frac{f'(c)}{g'(c)}$$  

In solving these type of question we need to consider a function. I want to know is there any specific method to find that function which is used to prove above type of equations or i have to learn its solution? Please explain the method if any.
 A: The idea is to use Rolle’s theorem, so we have to build a function $F$ such that applying to it Rolle’s theorem provides us with Cauchy’s MVT.
How can we prove Lagrange’s MVT from Rolle’s theorem? We consider $F(x)=f(x)-kx$ and choose $k$ so that the hypotheses of Rolle’s theorem apply: $F(a)=f(a)-ka$, $F(b)=f(b)-kb$, so we need
$$
f(b)-kb=f(a)-ka
$$
which means
$$
k=\frac{f(b)-f(a)}{b-a}
$$
Now the further step is easy: use $F(x)=f(x)-kg(x)$. We want
$$
f(a)-kg(a)=f(b)-kg(b)
$$
and therefore
$$
k=\frac{f(b)-f(a)}{g(b)-g(a)}
$$
Note that $g(b)\ne g(a)$ because, by assumption, the derivative $g'$ doesn't vanish over $(a,b)$.
Since Rolle’s theorem provides $c\in(a,b)$ where $F'(c)=0$, we have
$$
f'(c)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)=0
$$
or, equivalently,
$$
\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}
$$
The trick is essentially forming a linear combination of $f$ and $g$.
A: Hint: WLOG let $f(a)+g(a)\leq  f(b)+g(b)$ and then apply Roll theorem with
$$h(x)=\ln\Big[\big((f(x)-f(a)\big)\big(g(x)-g(b)\big)\Big]$$
on $[a,b]$.
A: I won't propose any proof of the statement, since it is clear that this would not be what you are looking for.
Shortly, the answer to your question can't be but negative. There is no general strategy to prove a statement: experience and efforts are the only ingredient you must add to your recipe!
Experience often offers standard approaches: after you have studied many proofs in your books, you should have a set of tools that may apply to your own statement.
Efforts cannot be dispensed with: proving a theorem is a difficult taks, and sometimes you must try several ideas before you isolate a working one.
In you example, the smart trick of the proof is exactly the definition of the good function. Proofs are a crative job, and you can't reduce them to a set of rigid rules.
