Question

If xy+6e^y = 6e, find the value of y" at the point where x = 0

Answer 1

Implicit differentiation:
\[xy+6e^y=6e\\
y+x~y'+6e^y~y'=0\\
y'(x+6e^y)=-y\\
y'=-\frac{y}{x+6e^y}\\
\large y''=-\frac{(x+6e^y)~y'-y\left(1+6e^y~y'\right)}{(x+6e^y)^2}\]
Now plug in the given number: \(x=0\)
\[\begin{align*}x=0~&\Rightarrow~y=1\\
&\Rightarrow~y'=-\frac{1}{0+6e}=-\frac{1}{6e}\\
&\large\Rightarrow~y''=-\frac{(0+6e)~\left(-\frac{1}{6e}\right)-\left(1+6e~\left(-\frac{1}{6e}\right)\right)}{(0+6e)^2}=\cdots
\end{align*}\]