text: let w = (siny)e^x , x=st, y=s+t. determine delw/delt and del2w/delsdelt at point (s,t)=(1,1)? the theorem is delw/delt= (delw/delx)/(delx/delt) + (delw/dely)/(dely/delt). but i have no idea how to get del2w/delsdelt this is a directional derivative and you have to have a direction to the derivative. any way let the direction be (th). then  w = f(y,x) and y = f(s,t)  and x = f(s,t)  note: dw/dt means the total derivative of w in the direction of (th) and it’s equals to  dw/dt = (pw/pt) sin(th) + (pw/ps) cos(th) ……. where p is partial derivative  also,  pw/pt = pw/py * py/pt + pw/px * px/pt……. where p is partial derivative  for  w= (siny) exp(x) …. and x = s t …… and y = s + t  then  pw/pt = (cos y) exp(x) * 1 + (sin y) exp(x) *s  and  pw/ps = pw/py * py/ps + pw/px * px/ps  pw/ps = (cosy) exp(x) * 1 + (siny) exp(x) *t  for dw/dt  dw/dt = [ (cos y) exp(x) * 1 + (sin y) exp(x) *s] sin(th) + [(cosy) exp(x) * 1 + (siny) exp(x) *t] cos(th)  now  d^2w/dsdt = d/ds[dw/dt]  then  d^2w/dsdt = d/ds{[ (cos y) exp(x) * 1 + (sin y) exp(x) *s] sin(th) + [(cosy) exp(x) * 1 + (siny) exp(x) *t] cos(th)}  d^2w/dsdt = [(siny) exp(x)] sin(th) ……….. eq(1)  for s =1 and t = 1 --->  x  = 1   and  y = 2  substitute for x and y in eq.1  d^2w/dsdt = [sin(2) exp(1)] sin(th)   and that is the second mix directional derivative of w at ( s =1 and t =1 ) in the direction of (th)