# Math Help - derivative check 2

1. ## derivative check 2

$
\frac{f(x)}{g(x)} = \frac{\sqrt{x^2+1}cosx}{x} = \frac{(x^2+1)^\frac{1}{2}cosx}{x}
$

$
f ' (x) = \frac{1}{2}(x^2+1)^\frac{-1}{2}cosx - \sqrt{x^2+1}sinx
$

so

$
y ' = \frac{x \frac{1}{2}(x^2+1)^\frac{-1}{2}cosx - \sqrt{x^2+1}sinx - \sqrt{x^2+1}cosx}{x^2}
$

correct? thank you

2. ## hi

Hi

I did not go through all your calculations, but I noticed something at the
beginning. Can you see what it is?

$\sqrt{x^2 + 1} = (x^2 + 1)^{1/2}$

You wrote: $\sqrt{x^2 + 1} = (x^2 + 1)^{-1/2}$

Can you see which one is correct?

3. Originally Posted by Twig
Hi

I did not go through all your calculations, but I noticed something at the
beginning. Can you see what it is?

$\sqrt{x^2 + 1} = (x^2 + 1)^{1/2}$

You wrote: $\sqrt{x^2 + 1} = (x^2 + 1)^{-1/2}$

Can you see which one is correct?
yes sorry mate i ment 1/2 not -1/2!! but wierd i didnt add it to the next line xD