# Simplify the expression

1. Simplify the derivative

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

3. $f'(x) = \frac{x*\frac{-2x}{2\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}}=\frac{-x^{2}-\sqrt{1-x^{2}}}{x^{2}\sqrt{1-x^{2}}}$ I just don't know how to simplify this further.

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SammyS
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1. Simplify the derivative

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

3. $f'(x) = \frac{x*\frac{-2x}{2\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}}=\frac{-x^{2}-\sqrt{1-x^{2}}}{x^{2}\sqrt{1-x^{2}}}$ I just don't know how to simplify this further.
You simplified $\displaystyle \ \fracx\frac{-2x}{2\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}} \$ incorrectly.

So it would be -x$^{2}$ over the square root. So how do I find the second derivative of $\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}}$ Do I use the product rule or quotient rule. Which is easier?

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SammyS
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Multiply by $\displaystyle \ \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \ .$

Multiply by $\displaystyle \ \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \ .$
I already did that and came out with $\frac{-1}{x^{2}\sqrt{1-x^{2}}}$

How do I take the second derivative?

haruspex
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How do I take the second derivative?
You can use the quotient rule. To differentiate the denominator use the product rule. Alternatively, rewrite it as a product of two terms with negative exponents and just use the product rule.

Here is what I have $\frac{\frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} x^{2} + \frac{2}{x^{3}}$ The squared on the top is supposed to go with the x on the bottom.

SammyS
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Here is what I have $\frac{\frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} x^{2} + \frac{2}{x^{3}}$ The squared on the top is supposed to go with the x on the bottom.
You mean $\displaystyle \ \ \frac\frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} {x^{2}} + \frac{2}{x^{3}}\ \$

I don't see how that can possibly be correct.

You mean $\displaystyle \ \ \frac\frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} {x^{2}} + \frac{2}{x^{3}}\ \$

I don't see how that can possibly be correct.

Here is how I got that answer $\frac{\frac{-2x^{3}}{2(1-x^{2})^{3/2}}}+\frac{2x}{2\sqrt{1-x^{2}}}-\frac{2x}{\sqrt{1-x^{2}}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}} = \frac{\frac{-2x^{3}}{2(1-x^{2})^{3/2}}}-\frac{x}{\sqrt{1-x^{2}}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}} = \frac{\frac{-x}{\sqrt{1-x^{2}}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}}=\frac{\frac{-x}{\sqrt{1-x^{2}}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}{x^{2}}+\frac{2}{x^{3}}$ That is how I got the answer. The + sign in the first part should be in the numerator and the x^2 should be in the denominator. Same thing with the sign in the second half of the first part and the sqrt should be in the numerator with the x^3 on the bottom, same thing in the second part and third part and last part. Sorry it got messed up.
It doesn't need to be that complicated. Write the first derivative as $-x^{-2}(1-x^2)^{-\frac 12}$ and differentiate using the product rule. Hint: every term in the answer ought to have a factor like $(1-x^2)^{n-\frac 12}$, some integer n. I'd guess that's how SammyS knew your answer could not be right.