How Do You Derive tan^(-1)(sqrt(8x^2-1))?

cemar.
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Derivative arctan function! Please help!

Find the derivative of tan^(-1)(sqrt(8x^2-1)))

I knwo that the derivative of tan^(-1)(x) is 1/(1+x^2) and that you are supposed to use the chain rule twice but i cannot seem to get the right answer.
If some one could please show me the steps i could figure out where i went wrong! Thanks!
 
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Show us what you did first. The first chain rule involves the tan^(-1) and the second involves the sqrt.
 
okay what i did was

1. f(x) = arctan((sqrt(8x^2-1)))

2. f'(x) = ((1/(1+(8x^2-1))) * (1/2)(8x^2-1)^(-1/2) * 16x

3. = 16x/((8x^2) * 2 * sqrt(8x^2-1))

4. = 1/(x(sqrt(8x^2-1))
 
I believe that is correct.
 
really?! That would explain why i was so confused. These online assignments are no fun. I guess i just have to work on submitting my answers properly. Ill just double check (again).
and thank you so much! I have been tearing my hair out over this for about an hour now trying to figure out where i went wrong.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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