Express $f'''_{xxx} and f'''_{yyy}$ in terms of $f'''_{uuu} and f'''_{vvv}$. Let $f(x,y)\in C^3(\mathbf{R}^2)$ and let $u=x+y$ and $v=y$.
Express $f'''_{xxx} and f'''_{yyy}$ in terms of $f'''_{uuu} and f'''_{vvv}$.
I'm supposed to use the chain rule, how do I go about?
Thanks!
Alexander
 A: Hint:
$$f'_x = f'_u u'_x + f'_v v'_x = f'_u $$
$$f''_x= (f'_x)'_x = (f'_u)'_x = (f'_u)'_u u'_x + (f'_u)'_v v'_x = f''_u$$
A: If $f$ is of the form $f(x_1,\ldots,x_n)$ I will use the notation $D_if$ for the partial derivativeof $f$ w.r.t. the $i$-th variable. Similarly, $D_{k,i}f=D_kD_if$ denotes the partial derivative of $D_if$ w.r.t. the $k$-th derivative.
Then $$\begin{align}D_1\left(f(x+y,y)\right)&=D_1f(x+y,y)D_1(x+y)+D_2f(x+y,y)D_1(y)\\&=D_1f(x+y,y)\end{align}$$
$$\begin{align} D_2 (f(x+y,y))&=D_1f(x+y,y)D_2(x+y)+D_2f(x+y,y)D_2(y)\\&=D_1f(x+y,y)+D_2f(x+y,y)\end{align}$$
Continuing, $$\begin{align} D_{1,1}(f(x+y,y))&=D_1(D_1(f(x+y,y))\\&=D_1(D_1f(x+y,y))\\&=D_{1,1}f(x+y,y)D_1(x+y)+D_{2,1}f(x+y,y)D_1(y)\\&=D_{1,1}f(x+y,y)\end{align}$$
$$\begin{align} D_{2,2}(f(x+y,y))&=D_2(D_2(f(x+y,y))\\&=D_2(D_1f(x+y,y))+D_2(D_1f(x+y,y))\\
&={{D_{1,1}}f\left( {x + y,y} \right){D_2}\left( {x + y} \right) + {D_{2,1}}f\left( {x + y,y} \right){D_2}\left( {x + y} \right)}\\
&+{D_{1,2}}f(x + y,y){D_2}\left( {x + y} \right) + {D_{2,2}}f(x + y,y){D_2}\left( y \right)\\&={D_{1,1}}f\left( {x + y,y} \right) + {D_{2,1}}f\left( {x + y,y} \right) + {D_{1,2}}f(x + y,y) + {D_{2,2}}f(x + y,y)\\&= {D_{1,1}}f\left( {x + y,y} \right) + 2{D_{1,2}}f(x + y,y) + {D_{2,2}}f(x + y,y)\end{align}$$
...and so on.
Note that $D_2(f(x+y,y))$ here denotes the partial derivative of the function composite function $f\circ g$, where $g(x,y)=(x+y,y)$, in parenthesis, while $D_2f(x+y,y)$ denotes the partial derivative $D_2f$ evaluated at $(x+y,y)$.
