Question

using implicit differentiation, find dy/dx of y=sin(xy)

Answer 1

y=sin(xy)
dy/dx(y)=dy/dx(sinxy)
y'=cosxy*dy/dx(xy)
y'=cosxy*(y+y'x)
y'/cosxy=y+y'x
y'/cosxy-y'x=y
y'(1/cosxy-1)=y
y'=(cosxy)/y

Answer 2

or actually just dy/dx=ycos(xy)

Answer 3

it says my answer is supposed to be ycos(xy)/1-xcos(xy)

Answer 4

@cljohn solution is good up to the 2nd to last line. When you factor out y' you forgot the x :)
\[y'(\frac{1}{cosxy}-x)=y\]

Answer 5

yeah I tried to show my work and solve at the same time on the computer, not helpful at all, Sorry :(

Answer 6

\[y'(\frac{1}{cosxy}-x)=y\]
\[y'(\frac{1}{\cos(xy)}-\frac{x\cos(xy)}{\cos(xy)})=y\]
\[y'\left(\frac{1-x\cos(xy)}{\cos(xy)}\right)=y\]
Now multiply both sides by \(\large \dfrac{\cos(xy)}{1-x\cos(xy)}\)

Answer 7

yup, thanks for helping out :)