$y=f(x) \in C^1$ is defined implicitly by $ax + by = f(x^2+y^2), f'(x) = ?$ Problem
$y=f(x) \in C^1$ is defined implicitly by $ax + by = f(x^2+y^2)$. $a$ and $b$ are constants . $f'(x) = ?$
Analysis
The answer for this exercise given by my teacher is $$f'(x) = \frac{2xf'(x^2+y^2)-a}{b-2yf'(x^2+y^2)}$$ I can't agree with that. Because if you express a derivative function with itself, in fact you don't find it.
Is it possible to write the expression of $f'(x)$ with only $x, y, a $ and $b$ ?
Thanks!
 A: $$\begin{cases}
y=f(x) \\
ax + by = f(x^2+y^2)
\end{cases}$$
In fact this is a functional equation :
$$ax + bf(x) = f(x^2+f(x)^2)$$
The unknown is the function $f$. But it is not asked to solve the functional equation, which seems very difficult.
The question appears ambiguous. From the given answer it seems that it is asked for the relationship between the derivatives $f'(x)$ and $f'(x^2+y^2)$. If this is the meaning of the question, a way to obtain the wanted relationship is shown below, using the differentials of the above functions.
$dy=df(x)=f'(x)dx$
$df(x^2+y^2)=f'(x^2+y^2)d(x^2+y^2)=f'(x^2+y^2)\big(2xdx+2ydy \big)$
$df(x^2+y^2)=f'(x^2+y^2)\big(2xdx+2yf'(x)dx \big)$
$df(x^2+y^2)=2f'(x^2+y^2)\big(x+yf'(x) \big)dx$
$$ax+by=f(x^2+y^2) \quad\implies\quad a_,dx+b\,dy=df(x^2+y^2)$$
$$a\,dx+b\,f'(x)dx=2f'(x^2+y^2)\big(x+yf'(x) \big)dx$$
$$a+bf'(x)=2f'(x^2+y^2)\big(x+yf'(x) \big)$$
$$f'(x)\big(b-2yf'(x^2+y^2) \big)=2f'(x^2+y^2)-a$$
$$f'(x) = \frac{2xf'(x^2+y^2)-a}{b-2yf'(x^2+y^2)}$$
Of course, this doesn't give the explicit derivatives $f'(x)$, but gives the relationship between the couples of derivatives at different values $(x)$ and $(x^2+y^2)$. This is probably what is expected because solving for explicit $f(x)$ or explicit $f'(x)$ is far to be an exercise of normal level for students.
