<s>[INST] Finding where the slope of tangent line is = 1 I am suppose to find the points on the curve $x^2+2y^2 = 1$ where the tangent line has a slope of 1.
I am not sure how to do this but I guessed that if I set it is
$y = \sqrt{\frac{1-x^2}{2}}$ then I can find that derivative and work from there.
I find the derivative as $y = (1-x^2)(\frac{1-x^2}{2})^\frac{3}{2}$
I am not sure how to work from here but just looking at the function it looks like x=1 would make it zero. I have absolutely no idea how to factor that, it just seems like a mess.
 [/INST] Don't solve for $y$ first; use implicit differentiation!
If $x^2+2y^2=1$, then
$$\begin{align*}
\frac{d}{dx}(x^2+2y^2) &= \frac{d}{dx} 1\\
2x + 4y\frac{dy}{dx} &= 0\\
4y\frac{dy}{dx} &= -2x\\
\frac{dy}{dx} &= -\frac{2x}{4y}\\
\frac{dy}{dx} &= -\frac{x}{2y}.
\end{align*}$$
If the tangent has slope $1$, then $\frac{dy}{dx}=1$, so
$$\begin{align*}
1 &= -\frac{x}{2y}\\
-2y &= x.
\end{align*}$$
Plugging into the original equation and solving for $y$, we get:
$$\begin{align*}
x^2 + 2y^2 &= 1\\
(-2y)^2 + 2y^2 &= 1\\
4y^2 + 2y^2 &= 1\\
6y^2 &= 1\\
y^2 &=\frac{1}{6}\\
|y| &=\sqrt{\frac{1}{6}}
\end{align*}$$
giving two values of $y$, and two corresponding values of $x$.
</s>