1. ## Derivative Help

Hello All,

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

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

Here is what I have so far...

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

$R=(4-x^2)$

$y=R^{1/2}$

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

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

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

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

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

Should I rationalize the denominator next?

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

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

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

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

2. No need to rationalize the denominator.

sqrt(4 - x^2) = -x
Square both sides. Simplify and solve for x.

3. $1=\frac{-x}{\sqrt{4-x^2}}$

$\sqrt{4-x^2}=-x$

$(\sqrt{4-x^2})^2=(-x)^2$

$4-x^2=x^2$

$4=2x^2$

$2=x^2$

$\pm\sqrt{2}=x$

Is this correct?

4. Originally Posted by dbakeg00
$1=\frac{-x}{\sqrt{4-x^2}}$

$\sqrt{4-x^2}=-x$

$(\sqrt{4-x^2})^2=(-x)^2$

$4-x^2=x^2$

$4=2x^2$

$2=x^2$

$\pm\sqrt{2}=x$

Is this correct?
Well, check it in the original equation! If $x= \sqrt{2}$ then $x^2= 2$ so $\sqrt{4- x^2}= \sqrt{4- 2}= \sqrt{2}$
$\frac{x}{\sqrt{4- x^2}}= \frac{\sqrt{2}}{\sqrt{2}}= 1$

On the other hand, if $x= -\sqrt{2}$, $\sqrt{4- x^2}$ is still $\sqrt{2}$ but now we have $\frac{x}{\sqrt{4- x^2}}= \frac{-\sqrt{2}}{\sqrt{2}}= -1$ which is not 1- but your original equation allowed " $\pm$".

Actually, your original equation had $y= \pm\sqrt{4- x^2}$ which is equivalent to $x^2+ y^2= 4$, a circle with center at (0,0) and radius 2. The tangent line will have slope 1 where that circle intersects the line with slope -1, y= -x. If y= -x, $y^2= x^2$ and $x^2+ y^2= x^2+ x^2= 2x^2= 4$ so $x^2= 2$ and $x= \pm\sqrt{2}$ but notice that the tangent line has slope 1 at $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2}$. The points $(\sqrt{2}, \sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$ also are on that circle, have $x= \pm\sqrt{2}$ but the tangent lines do not have slope 1 there.

You know, you really did not need to open a new thread just to continue
http://www.mathhelpforum.com/math-he...erivative.html

By the way, what is a "geographic" derivative?

5. By the way, what is a "geographic" derivative?
I meant geometric derivative...thats what the book I'm self studying from calls it, not my title. Sorry about the typo.

Thank you for the help. I haven't studied circles much, so equations like that are semi-foreign to me.