Question

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

Answer 1

no thats what i was given
implicit diff i think i just cant get the right answer

Answer 2

dy/dx is part of it?

Answer 3

no find the derivative of ln(xy) = e^(x+y)

Answer 4

thats better

Answer 5

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

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

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

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

Answer 6

ok cool tyvm

Answer 7

yep

Answer 8

i actually am having touble finding the dy/dx

Answer 9

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

Answer 10

on my homework it doesn't have e^x+y, I want to know if lnxy=x+y then find y'