Originally Posted by

**RedBarchetta** Find dy/dx:

$\displaystyle

\

\ln y = e^y \sin x

\

$

So now I simply take the derivative of both sides and solve for y':

$\displaystyle

\begin{gathered}

\frac{1}

{y}y' = y'e^y \sin x {\color{red}-} e^y \cos x \hfill \\

y' = y \cdot y'e^y \sin x - ye^y \cos x \hfill \\

ye^y \cos x = y \cdot y'e^y \sin x - y' \hfill \\

ye^y \cos x = y'(ye^y \sin x - 1) \hfill \\

y' = \frac{{ye^y \cos x}}

{{ye^y \sin x - 1}} \hfill \\

\end{gathered}

$

My answer book says different. I can't find a fault in my work. Here is what the book says:

$\displaystyle

y' = \frac{{ye^y \cos x}}

{{1 - ye^y \sin x}}

$

Any help will be deeply appreciated. Thank you!