$\displaystyle e^y+24-e^-1 = 5x^2+4y^2 $ I can't figure this out.... I believe this stage is correct $\displaystyle e^y+e^-1=10x+8y*y'-y'$ But Need help on further solving it. Help appreciate.
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Actually the first step should be $\displaystyle y'e^y = 10x + 8yy'$
Where did you put the $\displaystyle e^-1$?
$\displaystyle e^{-1}$ is a constant and its derivative is 0.
I thought it was -e^-1. Anyway, in case of constant I get $\displaystyle (10x)/(-1*e^y-8y)$. Where could I have gone wrong there?
I get $\displaystyle y' = \frac{10x}{e^y - 8y}$
The reason i have negative 1 is because there are two y', and when i place it ont he other side (-y') I factored and assumed 1. Why is this not correct? Thanks however.
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