Given the function $f: \Bbb R^2 \to \Bbb R$ has continuous partial derivatives at $(1,1)$ and it is also known that $f_x(1,1)=-1$ , $f_y(1,1)=2$ and $f(1,1)=1$
calculate $g(x)=f(x,f(x,x))$
According to the solution in the book the answer is $g'(x)=f_x+f_y \cdot (f_y+f_x)=1$
I thought it should be $f_x \cdot x' + f_y \cdot 0 $ since we are doing derivative with respect to $y$ and here we have $x$?
Edit-
what they did is say $(x,y)=(t,f(t,t))$ and then $g'(t)=f_x \cdot \frac{dx}{dt} + f_y \cdot \frac{dy}{dt}$ now in our case $\frac{dy}{dt}$ is $f'(x,x)$ But $f'(x,x)$ with respect to what? why in the solution it is written as $f_y+f_x$? this is what I cannot understand
Thanks for any help and tips
The notation is overloading the symbol $x$ in a way that leads to confusion for the unwary.
They are using $f_x$ as the derivative of $f$ with respect to its first argument, and $f_y$ as the derivative of $f$ with respect to its second argument. Where as elsewhere $x$ is used as a free variable.
This is poor practice, and why the suggestion was to use $t$ as the symbol for the free variable, so you can then find $g'(1)$ - the derivative of $g$ evaluated at $1$.
$$\begin{align}g(t) &= f(t, f(t,t))\\[1ex]g'(t) & = f_x(t,f(t,t))+f_y(t,f(t,t))\cdot(f_x(t,t)+f_y(t,t))\\[1ex]g'(1) & = f_x(1,f(1,1))+f_y(1,f(1,1))\cdot(f_x(1,1)+f_y(1,1))\\&=f_x(1,1)+f_y(1,1)\cdot(f_x(1,1)+f_y(1,1))\\&=-1+2\cdot(-1+2)\\&=1\end{align}$$
Alternatively you might use $\partial_1 f(~,~)$ and $\partial_2 f(~,~)$ to indicate derivatives with respect to the first and second arguments of the function.
$$\begin{align}g(x) &= f(x, f(x,x))\\ g'(x) & = \partial_1 f(x,f(x,x))+\partial_2 f(x,f(x,x))\cdot(\partial_1 f(x,x)+\partial_2 f(x,x))\\[1ex]g'(1) & = \partial_1 f(1,f(1,1))+\partial_2 f(1,f(1,1))\cdot(\partial_1 f(1,1)+\partial_2 f(1,1))\\&= \partial_1 f(1,1)+\partial_2 f(1,1)\cdotp(\partial_1 f(1,1)+\partial_2 f(1,1))\\&=1\end{align}$$