Question

Suppose g(r,s)=f(2r-r, s^2-4r) and fx(0,0)=2, fy(0,0)=-4. Find gr(1,2) and gs(1,2).

Answer 1

i think \(g(r,s)=f(2r-s, s^2-4r) \) ???

Answer 2

suppose that \(z=f(x,y)\) is a function of \(x\) and \(y\) and let \(x=x(r,s)\) , \(y=y(r,s)\)

( \(x\) , \(y\) are functions of \(r\) and \(s\) )

Chain Rule for Functions of Two Variables says

\[\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \]
and
\[\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \]

Answer 3

can you show me how to do the first set of partials.. i dont understand how to get the partialz/partialx

Answer 4

well there is no relation given for \(z=f(x,y)\) but note that \(f_x(0,0)=2, f_y(0,0)=-4\)

Answer 5

for this case \(x=2r-s \) and \(y=s^2-4r \) so note that for \((r,s)=(1,2) \) u have \((x,y)=(0,0)\)

Answer 6

make sense?

Answer 7

i mean u cant derive a relation for partialz/partialx (there is no realtion given for z in terms of x and y) but actually u have the value of partialz/partialx(0,0)