Question

xe^y=x-y

Answer 1

xe^y=x-y
e^y = (x-y)/x
y = ln ((x-y)/x)

Answer 2

What do you have to do here?

Answer 3

hello Professor @SolomonZelman :)

Answer 4

don't get fooled by the title... I'm not....

Answer 5

@awalton63 , can you give us more clear instructions?

Answer 6

Find dy/dx by implicit differentiation

Answer 7

I would do this:
\(\large\color{black}{ xe^y=x-y}\)

\(\large\color{black}{ e^y=\frac{\LARGE {x-y} }{\LARGE {x}}}\)

\(\large\color{black}{\ln(e^y)=\ln(\frac{\LARGE {x-y} }{\LARGE {x}})}\)

\(\large\color{black}{\ln(e^y)=\ln(x-y )-\ln(x)}\)

Answer 8

Now take out the y from the exponent, and you get
\(\large\color{black}{y\ln(e)=\ln(x-y )-\ln(x)}\)

Answer 9

Then take the derivative.

Answer 10

let me know if you need more help, with a concept or with proving it.
Ask any question about the problem.

(Don't forget the chain rule for the y, since you are finding dy/dx)

Answer 11

@awalton , I would prefer you to post you work here, so that I can see the mistake if you have any.... okay?

Answer 12

You seem not to reply. Do you have any problems with the problem? :))

Answer 13

xe^y=x-y
x(e^y)'+x'(e^y)=x'-y'
xe^y+e^y=1-y'

Answer 14

i originally was trying to isolate the y to one side which is where I ran into trouble and I came up with x-x=-y/e^y

Answer 15

Just derive it the way it is.
\(\large\color{black}{y\ln(e) =\ln(x-y)-\ln(x) }\)

Answer 16

u can do that too :)
but its better to use differentiate properties :)
anyway all good :P

Answer 17

The derivative of ln(x) is 1/x.

The proof
\(\large\color{black}{x =\ln y }\)
\(\large\color{black}{x =\log_e y }\)
\(\large\color{black}{e^y =x }\)
\(\large\color{black}{\frac{dy}{dx}e^y =1 }\)
\(\large\color{black}{\frac{dy}{dx} =1/e^y }\)
sub in x for e^y since you know that x=e^y
\(\large\color{black}{\frac{dy}{dx} =1/x }\)

Answer 18

the ln(x-y) will be as follows:

Apply the rule to derive "ln(x-y)" for the inner argument, the x-y,
and by the chain rule, multiply times the derivative of (x-y)
and another chain will apply to the derivative of y ....

Answer 19

and the derivative of yln(e), will be just ln(e) times dy/dx (dy/dx comes from the chain rule for y)

Answer 20

let me know if you need more assistance.