For $e^y + 19 - e^{-1} = 4 x^2 + 3 y^2$, evaluate $y'$ at the point $(-2,-1)$.

Question

For the equation given below, evaluate $y'$ at the point $(-2,-1)$. $$e^y + 19 - e^{-1} = 4 x^2 + 3 y^2$$

\begin{align}e^y + 19 - e^{-1} &= 4 x^2 + 3 y^2\\ e^y + 0 - 0 &= 8x + 6y \cdot y'\\\\ \frac{e^y - 8x}{6y} &= y'\\\\ \frac{e^{-1} - 8(-2)}{6(-1)} &= y'\\\\ \frac{16.36787944}{-6} &= y'\end{align}

I don't understand what I'm doing wrong?

Answer 1

$e^y + 19 - e^{-1} = 4x^2 + 3y^2$.

$e^yy' = 8x+ 6yy'$.

$y'(e^y-6y)=8x$.

$\displaystyle y' = \frac{8x}{e^y-6y}$.

Evaluated at $(-2,-1)$ gives us $\displaystyle \frac{-16}{e^{-1}+6}=\frac{dy}{dx}=\boxed{-2.513}$

You just forgot that $\displaystyle \frac{d}{dx}e^y = e^yy'$.