Let $a,b$ be constants satisfying $a^2+b^2=1.$ Let $f=f(x,y)$ be a twice differentiable function, and $u=ax-by$ and $v=bx+ay$. For the function $g$ defined by $g(x,y)=f(u,v)$, prove $g_{xx}(x,y)+g_{yy}(x,y)=f_{uu}(u,v)+f_{vv}(u,v)$
i was studying for my exams and i got stuck. I'm not sure what I'm supposed to do here... I can't even start.... a little help would be much appreciated :'(
Use the chain rule twice:
$$g_x = f_u u_x + f_v v_x= af_u + bf_v$$
$$g_{xx} = a(af_u + bf_v)_u + b(af_u + bf_v)_v = a^2f_{uu}+abf_{uv}+abf_{vu}+ b^2f_{vv}$$
$$g_y = f_u u_y + f_v v_y= -bf_u + af_v$$
$$g_{yy} = -b(-bf_u + af_v)_u + a(-bf_u + af_v)_v = b^2f_{uu}-abf_{vu}-abf_{uv}+ a^2f_{vv}$$
Now add.