Please help again

[thelazyman]

:lol:


Please help me with this question:

For each of the following, determine the equation of the tangent at the given point.

How do you isolate y in this case

for the curve defined by x^2e^y at B(0,1)

Please help

 

[galactus]

Maybe they want implicit differentiation

 

[soroban]

for the curve defined by x^2e^y

You didn't give us an equation . . .

 

[thelazyman]

x^2e^y=0 at B(0,1) is the equation. Sorry

 

[galactus]

Try implicit differentiation.

\(\displaystyle x^{2}e^{y}\frac{dy}{dx}+(2x)e^{y}=0\)

\(\displaystyle x^{2}e^{y}\frac{dy}{dx}=-(2x)e^{y}\)

\(\displaystyle \frac{dy}{dx}=\frac{-2xe^{y}}{x^{2}e^{y}}\)

\(\displaystyle \frac{dy}{dx}=\frac{-2}{x}\)

Now, given your coordinates, do you see a problem when you enter x=0 into your equation?.

 

[Unco]

Thelazyman:

We can make \(\displaystyle \mbox{y}\) the subject of
\(\displaystyle \mbox{ x^2 \cdot e^y = 1\)

Divide both sides by \(\displaystyle \mbox{x^2}\):

\(\displaystyle \mbox{ e^y = \frac{1}{x^2}}\)

Take natural logs of both sides and simplify:

\(\displaystyle \begin{align*}
\mbox{ y } &= \mbox{ \ln{\left(\frac{1}{x^2})}} \\
\\
\\
&= \mbox{\ln{(x^{-2})}} \\
\\
\\
&=\mbox{ -2\ln{x}}\\
\end{align*}\)

You could differentiate from here (explicitly):

\(\displaystyle \L \mbox{ \frac{dy}{dx} = \frac{-2}{x}}\)