Question

Find dy/dx using implicit differentiation: xe^y=y-1

Answer 1

..imply it then :) let y be some function of x, and apply the chain rule to the usual derivative rules

Answer 2

i see the basics: product rule, e rule, power rule and constant rule

Answer 3

so d/dx (xe^y) = d/dx (y-1) to start it off?

Answer 4

yep, thats a fine start if you like that kind of notation

Answer 5

I know d/dx = y-1 = ydx/dy im kind of stuck on d/dx(xe^y)

Answer 6

d/dx (y - 1)

d/dx (y) - d/dx (1)

dy/dx - 0

Answer 7

d/dx (xe^y) ; is a product rule at heart

d/dx (x) e^y + x d/dx (e^y)

dx/dx e^y + x dy/dx e^y

e^y + x dy/dx e^y

Answer 8

if you view this as the d/dx "absorbing" the variable; it gets easier to correlate

Answer 9

\[\frac{d}{dx}(e^y)\]
absorb the y, and run a normal e^x outcome
\[\frac{dy}{dx}~e^y\]

Answer 10

Oh so i got e^y/1-xe^y

Answer 11

maybe, let me try it my way:

\[x~ e^y = y - 1\]
\[e^y+x~ y'e^y = y'\]
\[e^y = y'-x~y'e^y\]
\[e^y = y'(1-x~e^y)\]
\[y'=\frac{e^y}{1-x~e^y}\]

Answer 12

You can also think of d/dx(e^y) as d/dy(e^y)dy/dx = (e^y)(dy/dx)

Answer 13

looks good to me

Answer 14

thanks a lot

Answer 15

good luck

Answer 16

ranga, i cant read thru that notation to well to comment :)

Answer 17

i think i see it now ...

Answer 18

\[\frac{ d }{ dx}e ^{y} = \frac{ d }{ dy }e ^{y}\frac{ dy }{ dx } = e ^{y}\frac{ dy }{ dx }\]