Calculate all four second-order partial derivatives of f(x,y)=sin(3x/y). I got fxx but I can't get the ones that deals with y's. fxx=(-9/y^2)sin(3x/y) fxy= fyx= fyy=

Question

amistre64 · Answer

fxy means fx then fy the results

amistre64 · Answer

$$f(x,y)=sin(3\frac{x}{y})$$
$$fx=\frac{3}{y}cos(3\frac{x}{y})$$
$$fxx=-\frac{3}{y}\frac{3}{y}sin(3\frac{x}{y})$$

amistre64 · Answer

$$fx=\frac{3}{y}cos(3\frac{x}{y})$$
$$fxy=\frac{3}{y}'cos(3xy^{-1})+\frac{3}{y}cos'(3xy^{-1})$$
might be more helpful
$$fxy=\frac{3}{y}'cos(3xy^{-1})+\frac{3}{y}cos'(3xy^{-1})$$

amistre64 · Answer

$$fxy=\frac{3}{y}'cos(3xy^{-1})+\frac{3}{y}cos'(3xy^{-1})$$
$$fxy=-\frac{3}{y^2}cos(3xy^{-1})+\frac{3}{y}\frac{3x}{y^{2}}sin(3xy^{-1})$$
maybe

amistre64 · Answer

it tricky, but if you keep your wits about you its doable

anonymous · Answer

y is the first part in fxy cos?