1. ## Simple derivative

Hi I need to find the derivative of $\displaystyle \frac{1}{\sqrt{1-x^2}}$.I found it as:
$\displaystyle \frac{d}{dx} (\frac{1}{\sqrt{x^{2}-1}})\newline$
$\displaystyle \frac{d}{dx} (x^{2}-1)^{-\frac{1}{2}}=-\frac{1}{2}(x^{2}-1)^{-\frac{1}{2}-1} \newline$
$\displaystyle =-\frac{1}{2(x^2-1)^\frac{3}{2}}$

But the answer by the book is $\displaystyle -\frac{x}{(x^2-1)^\frac{3}{2}}$

Help.

2. ## Re: Simple derivative

Your function is \displaystyle \displaystyle \begin{align*} y = \left( x^2 - 1 \right) ^{-\frac{1}{2}} \end{align*}, which is a COMPOSITION of functions. So to differentiate this you will need to use the chain rule.

You have done the outer derivative correctly, but you have forgotten that you also need to find the inner derivative, which is \displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left( x^2 - 1 \right) = 2x \end{align*}. Multiply this with the result you got.

3. ## Re: Simple derivative

Well I have not studied chain rule yet.thanks for the help.

4. ## Re: Simple derivative

I'd be interested to know how to do this problem without the chain rule. Does anybody have any ideas?

- Hollywood

5. ## Re: Simple derivative

I can't see why you would want not to use the chain rule.

By the way, did you notice that your original function was $\displaystyle \frac{1}{\sqrt{1- x^2}}$ while the function you tried to integrate was $\displaystyle \frac{1}{\sqrt{x^2- 1}}$. Those are not at all the same thing!

6. ## Re: Simple derivative

He said he hadn't studied the chain rule yet, and I couldn't see a non-chain-rule solution. So I was curious.

I didn't notice the change in the function.

- Hollywood

7. ## Re: Simple derivative

Hi,
Since you haven't studied the chain rule yet, I assume you're just starting derivatives. So perhaps you want to find the derivative by definition? The attachment shows how to do this.