Question

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=

Answer 1

fxy means fx then fy the results

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

y is the first part in fxy cos?