Hello All,

Here is my equation:$\displaystyle y=\pm \sqrt{4-x^2}$

I'm supposed to find the value of x where the slope = 1

Here is what I have so far...

$\displaystyle y=(4-x^2)^{1/2}$

$\displaystyle R=(4-x^2)$

$\displaystyle y=R^{1/2}$

$\displaystyle \frac{dy}{dR}=\frac{1}{2\sqrt{4-x^2}}$

$\displaystyle \frac{dR}{dx}=-2x$

$\displaystyle \frac{dy}{dx}=\frac{-2x}{2\sqrt{4-x^2}}$

$\displaystyle \frac{dy}{dx}=\frac{-x}{\sqrt{4-x^2}}$

$\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$

Should I rationalize the denominator next?

$\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$

$\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$ X $\displaystyle \frac{\sqrt{4-x^2}}{\sqrt{4-x^2}}$

$\displaystyle 1=\frac{-x\sqrt{4-x^2}}{4-x^2}$

This is where I am stuck. Can anyone give me a tip on where to go next?