1. ## Exponential Function Derivative

Find dy/dx:

$
\
\ln y = e^y \sin x
\
$

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

$
\begin{gathered}
\frac{1}
{y}y' = y'e^y \sin x - 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:

$
y' = \frac{{ye^y \cos x}}
{{1 - ye^y \sin x}}
$

Any help will be deeply appreciated. Thank you!

2. Originally Posted by RedBarchetta
Find dy/dx:

$
\
\ln y = e^y \sin x
\
$

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

$
\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:

$
y' = \frac{{ye^y \cos x}}
{{1 - ye^y \sin x}}
$

Any help will be deeply appreciated. Thank you!
The fault is in the first line. The negative in red should be a positive. The derivative of sin x is cos x, NOT - cos x. This mistake has a ripple effect, easily fixed, which I'll let you work through.

3. Oops! There we go. Don't you hate it when you mix up integrating and differentiating?

Thanks Mr. Fantastic.