value of $\frac{d}{dx}f(x,f(x,x))$ $f(x,y)$ is differentiable at $(1,1)$, and $f(1,1)=\frac{\partial f}{\partial x}(1,1)=\frac{\partial f}{\partial y}(1,1)$, then what is the value of $\frac{d}{dx}f(x,f(x,x))$ when $x=1$?
I applied the definition of derivative but don't know how to proceed.
 A: Because the letter $x$ is used for either the first or the second variable, maybe it is best to denote the partial derivatives of $f$ using subscripts and not letters $x,y$:
$$
f_1(a,b) = \frac{\partial f(t,b)}{\partial t}\big\vert_{t=a}
\\
f_2(a,b) = \frac{\partial f(a,t)}{\partial t}\big\vert_{t=b}
$$
Then your answer is:
$$
\frac{d}{dx}f\big(x,f(x,x)\big) =
f_1\big(x,f(x,x)\big)+
f_2\big(x,f(x,x)\big)f_1(x,x)+
f_2\big(x,f(x,x)\big)f_2(x,x) .
$$
and
$$
\frac{d}{dx}f\big(x,f(x,x)\big)\big\vert_{x=1} =
f_1\big(1,f(1,1)\big)+
f_2\big(1,f(1,1)\big)f_1(1,1)+
f_2\big(1,f(1,1)\big)f_2(1,1) .
$$
A: Essentially you're substituting $x=x$ and $y=f(x,x)$; it's easier to break down the problem a little bit: let $g(x)=f(x,x)$ and $h(x)=f(x,f(x,x))=f(x,g(x))$. Then by the chain rule,
$$\dfrac{dh}{dx} = \dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} \dfrac{dg}{dx} = \dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} \left( \frac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} \right)$$
A: Write $f_x = \frac{\partial f}{\partial x}$ and $f_y = \frac{\partial f}{\partial y}$. Then:
\begin{equation*}
\frac{d}{dx} f(x, f(x,x)) = f_x \cdot \frac{dx}{dx} + f_y \cdot \frac{d}{dx} f(x,x) =
\end{equation*}
\begin{equation*}
= f_x + f_y \cdot \Big[ f_x \cdot \frac{dx}{dx} + f_y \cdot \frac{dx}{dx} \Big] =
\end{equation*}
\begin{equation*}
= f_x + f_x f_y + f_y^2
\end{equation*}
A: $$ = \frac{\partial}{\partial x}f(1,f(1,1))+ \frac{\partial }{\partial y} f(1,f(1,1)) \times
\left( \frac{\partial}{\partial x}f(1,1) +  \frac{\partial }{\partial y} f(1,1)\right)                    $$
