Question

e^(x+y) = x+y. find dy/dx

Answer 1

Take natural logarithm on both sides then take derivative of both sides then isolate y'.

Answer 2

so ln(e^x+y) (1/1+y') =ln(x+y) (1/1+y')?

Answer 3

doesn't the y' cancel out though?

Answer 4

im confused as to how to isolate y'

Answer 5

No, do natural log before derivative.

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

x+y = ln(x+y)

Answer 6

Now take derivative.

Answer 7

ok so then 1+y' = (1 +y')/(x+y)?

Answer 8

Yes!

Answer 9

Now try to isolate y'.

Answer 10

but then dont you get (x+y) = (1+ y')/(1+y')?

Answer 11

and thats cancels out then doesnt it?

Answer 12

im just confused as to how to isolate it

Answer 13

are you still there?

Answer 14

If I was to do this - I would take the natural log of both sides...
x+y = ln(x+y)

i would then leave the answer as y = ln(xy) - x

Answer 15

am i missing some of the directions?

Answer 16

Or simply y=-x

Answer 17

sorry, that's wrong. y=1-x

Answer 18

i will agree to that also, but I do not see why you need to take another derivative

Answer 19

i cannot have a y in my answer...i just submitted it and it says that y is not a variable in that context

Answer 20

how did you get y=1-x?

Answer 21

Okay, differentiate both sides then. e^(u)*d/dx(u)=1+1?

Answer 22

but if you differentiate y doesnt that become y' (or dy/dx) not 1?

Answer 23

yeah, I suppose, hmmm