# Thread: chain rule - second order - quick review

1. ## chain rule - second order - quick review

Let: $\displaystyle f(u,v)=usin(v)$

where: $\displaystyle u(x,y)=x-y$ and $\displaystyle v(x,y)=e^{y-x}$

Use only the chain rule to find $\displaystyle f_x, f_y, f_{xx}, f_{yy}$ in terms of v and u.

So,

$\displaystyle f_u*u_x+f_v*v_x$

$\displaystyle =sinv-e^{y-x}ucosv=sinv-vucosv$

$\displaystyle f_y=f_u*u_y+f_v*v_y$

$\displaystyle =-sinv+e^{y-x}ucosv=sinv+vucosv$

Correct?

Then:

Mostly wondering if this is the correct formula:

$\displaystyle f_{xx}=f_{uu}*u_{xx}+f_{vv}*v_{xx}$

In which case:

$\displaystyle f_{xx}=f_{yy}=(-usinv)e^{y-x}=-vusinv$

$\displaystyle f_{xy}=(-usinv)(-e^{y-x})=vusinv$

2. Originally Posted by billym
Let: $\displaystyle f(u,v)=usin(v)$

where: $\displaystyle u(x,y)=x-y$ and $\displaystyle v(x,y)=e^{y-x}$

Use only the chain rule to find $\displaystyle f_x, f_y, f_{xx}, f_{yy}$ in terms of v and u.

So,

$\displaystyle f_u*u_x+f_v*v_x$

$\displaystyle =sinv-e^{y-x}ucosv=sinv-vucosv$

$\displaystyle f_y=f_u*u_y+f_v*v_y$

$\displaystyle =-sinv+e^{y-x}ucosv=sinv+vucosv$

Correct?
So far, so good!

Then:

Mostly wondering if this is the correct formula:

$\displaystyle f_{xx}=f_{uu}*u_{xx}+f_{vv}*v_{xx}$
No. $\displaystyle f_{xx}= (f_x)_x= (f_u u_x+ f_v v_x)_x= (f_u)_x u_x+ f_u u_xx+ (f_v)_x v_x+ f_v v_{xx}$$\displaystyle = (f_{uu} u_x+ f_{uv} v_x)u_x+ f_u u_{xx}+ (f_{vu} u_x+ f_{vv} v_x)v_x+ f_v v_{xx}$.

In which case:

$\displaystyle f_{xx}=f_{yy}=(-usinv)e^{y-x}=-vusinv$

$\displaystyle f_{xy}=(-usinv)(-e^{y-x})=vusinv$

3. $\displaystyle f_u=sin(v), f_{uu}=0, f_v=ucos(v), f_{vv}=-usin(v), f_{uv}=f_{vu}=cos(v)$

$\displaystyle u_x=1, u_{xx}=0, v_x=-v, v_{xx}=v$

$\displaystyle u_y=-1, u_{yy}=0, v_y=v_{yy}=v, u_{xy}=0, v_{xy}=-v$

Have I made some calculation errors or are these totally wrong?

$\displaystyle f_{xx}=v[u(vsinv+cosv)-2cosv]$

$\displaystyle f_{yy}=v[u(cosv-sinv)-2cosv]$

$\displaystyle f_{xy}=-v[u(cosv-vsinv)-2cosv]$

(just a little tedious)