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.
 A: Your approach should work. We have $y^2=(1/2)(1-x^2)$, and therefore 
$$y =\pm \frac{1}{\sqrt{2}}\sqrt{1-x^2}.$$
The plus sign is for the top half of the ellipse, and the minus sign is for the bottom half. If you just stick with the top half, you will only find one point at which the tangent line has slope $1$.  (A quick sketch shows this, and tells you roughly what the answers should be, useful as a check later.)
Let's work with $y =\pm \frac{1}{\sqrt{2}}\sqrt{1-x^2}$, we can deal with the other one later. Now we want  to differentiate. By the Chain Rule, we have
$$\frac{dy}{dx}=\frac{1}{\sqrt{2}}(-2x)(1/2)(1-x^2)^{-1/2}=-\frac{x}{\sqrt{2}}(1-x^2)^{-1/2}.$$
We want the derivative to be $1$. So we want to solve the equation
$$-\frac{x}{\sqrt{2}}(1-x^2)^{-1/2}=1.$$
Note that $x$ needs to be negative. Now square both sides and simplify. You should get something like $x^2=2(1-x^2)$. Solve for $x$, remembering it is negative.
I leave it to you to deal with the other half of the ellipse.
There is a much easier approach, through implicit differentiation. Look at the relationship $x^2+2y^2=1$. Nice formula, don't want to mess it up by taking square roots. Differentiate with respect to $x$ immediately. We get
$$2x+4y\frac{dy}{dx}=0.$$
The slope of the tangent line is $1$, so $\frac{dy}{dx}=1$, and therefore from $2x+4y\frac{dy}{dx}=0$ we conclude that $2x+4y=0$, meaning that $x+2y=0$. But also $x^2+2y^2=1$. Substitute $-2y$ for $x$. We get $6y^2=1$. So $y=\pm\frac{1}{\sqrt{6}}$ and the corresponding $x$ are $x=\mp \frac{2}{\sqrt{6}}$.
A: Another approach would be the following: Let $f(x,y)=x^2+2y^2-1$. The tangent slope is given by: $$\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}=1$$   See (here for proof). Using this equation and $f(x,y)=0$ you can find the points where the slope is equal to one. 
A: 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 $x$.
