Expert: David Hemmer - 9/27/2010

Question
Need to find dy/dx  this is not a homework problem
ln y = (e^y)sin(x)
I took the Ln of both sides to get Ln(lny)=y + ln sinx.then I differentiated:
(1/lny)(1/y)dy=dy + cos x/sinx dx
dy/dx((1/lny)(1/y)- 1)= cosx/sinx if I just solve for dy/dx I will have functions of y on the right hand side. what do i do next?

Answer
You can do this without talking ln first:

(1/y) y' = e^y*y'*sin(x) + e^ycos(x)

y'=y(e^y*y'*sin(x) + e^ycos(x))

You shouldn't expect to get your answer just in terms of x, since the original equation can't necessarily be solved for y as a function of x. Instead you just get an equation for dy/dx in terms of both x and y. You can still use this to compute dy/dx and any particular given point on the graph.