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Math Help - Exponential Function Derivative

  1. #1
    Member RedBarchetta's Avatar
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    Exponential Function Derivative

    Find dy/dx:

    <br />
\<br />
\ln y = e^y \sin x<br />
\<br />

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

    <br />
\begin{gathered}<br />
  \frac{1}<br />
{y}y' = y'e^y \sin x - e^y \cos x \hfill \\<br />
  y' = y \cdot y'e^y \sin x - ye^y \cos x \hfill \\<br />
  ye^y \cos x = y \cdot y'e^y \sin x - y' \hfill \\<br />
  ye^y \cos x = y'(ye^y \sin x - 1) \hfill \\<br />
  y' = \frac{{ye^y \cos x}}<br />
{{ye^y \sin x - 1}} \hfill \\ <br />
\end{gathered} <br />

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

    <br />
y' = \frac{{ye^y \cos x}}<br />
{{1 - ye^y \sin x}}<br />

    Any help will be deeply appreciated. Thank you!
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Find dy/dx:

    <br />
\<br />
\ln y = e^y \sin x<br />
\<br />

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

    <br />
\begin{gathered}<br />
\frac{1}<br />
{y}y' = y'e^y \sin x {\color{red}-} e^y \cos x \hfill \\<br />
y' = y \cdot y'e^y \sin x - ye^y \cos x \hfill \\<br />
ye^y \cos x = y \cdot y'e^y \sin x - y' \hfill \\<br />
ye^y \cos x = y'(ye^y \sin x - 1) \hfill \\<br />
y' = \frac{{ye^y \cos x}}<br />
{{ye^y \sin x - 1}} \hfill \\ <br />
\end{gathered} <br />

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

    <br />
y' = \frac{{ye^y \cos x}}<br />
{{1 - ye^y \sin x}}<br />

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

    Thanks Mr. Fantastic.
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