Question

find dy/dx of ln(xy) = e^(x+y)

Answer 1

you have asked this before, do implicit differentiation

Answer 2

i tired but it will not accept the answer. and whenever i work it out i get something different

Answer 3

its not taking that either

Answer 4

ln(xy) = e^(x+y)


d/dx[ln(xy)] = d/dx[e^(x+y)]


(y+xy')/(xy) = d/dx[e^(x+y)]


(y+xy')/(xy) = (1+y')e^(x+y)


y+xy' = xy(1+y')e^(x+y)


y+xy' = xye^(x+y)+xyy'e^(x+y)


y - xye^(x+y) =xyy'e^(x+y)-xy'


y - xye^(x+y) =y'(xye^(x+y)-x)


(y - xye^(x+y) )/(xye^(x+y)-x) =y'


y' = (y - xye^(x+y) )/(xye^(x+y)-x)

Answer 5

Sorry I was mistaken :D he is right

Answer 6

bit of a mess, but you should be able follow (somewhat...lol)

Answer 7

so why when you do that you get (y+xyprime)/xy at the begining?

Answer 8

OHHH

Answer 9

got it
lol

Answer 10

tyvm that helped alot

Answer 11

d/dx[ln(xy)]


[(xy)']/(xy) .... derive ln(xy)...don't forget to use the chain rule


(y+xy')/(xy) .... use the product rule to derive (xy)

Answer 12

np