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

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

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

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.

haruspex
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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.