# Thread: Exponential Function Derivative

1. ## Exponential Function Derivative

Find dy/dx:

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

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

$\displaystyle \begin{gathered} \frac{1} {y}y' = y'e^y \sin x - e^y \cos x \hfill \\ y' = y \cdot y'e^y \sin x - ye^y \cos x \hfill \\ ye^y \cos x = y \cdot y'e^y \sin x - y' \hfill \\ ye^y \cos x = y'(ye^y \sin x - 1) \hfill \\ y' = \frac{{ye^y \cos x}} {{ye^y \sin x - 1}} \hfill \\ \end{gathered}$

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!

2. 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!
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.