# Thread: help with a derivative involving arctan

1. ## help with a derivative involving arctan

find derivative of
f(x)=(sq.rt.)[x] - arctan(sq.rt.)[x]

my book says ((sq.rt.)(x))/(2(1+x)
however, online i found it to be (1/(2(sq.rt.)[x])) - (1/(2(sq.rt.)[x](x-1)))

i tried to simplify either one into the other, but no luck, so that means one of them is wrong...however, i cant find the derivative initially, so it doesnt matter

i would appreciate it if anybody could help me solve this

2. Originally Posted by twostep08
find derivative of
f(x)=(sq.rt.)[x] - arctan(sq.rt.)[x]

my book says ((sq.rt.)(x))/(2(1+x)
however, online i found it to be (1/(2(sq.rt.)[x])) - (1/(2(sq.rt.)[x](x-1)))

i tried to simplify either one into the other, but no luck, so that means one of them is wrong...however, i cant find the derivative initially, so it doesnt matter

i would appreciate it if anybody could help me solve this
$f(x) = \sqrt{x} - \arctan(\sqrt{x})$

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

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

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

$f'(x) = \frac{x}{2\sqrt{x}(1+x)} = \frac{\sqrt{x}}{2(1+x)}$

3. makes sense, however howd you get derivative of arctan (sq.rt.[x]) to be
$
$
$\frac{\frac{1}{2\sqrt{x}}}{1 + x}
" alt="\frac{\frac{1}{2\sqrt{x}}}{1 + x}
" />

4. Originally Posted by twostep08
makes sense, however howd you get derivative of arctan (sq.rt.[x]) to be
$
$
$\frac{\frac{1}{2\sqrt{x}}}{1 + x}
" alt="\frac{\frac{1}{2\sqrt{x}}}{1 + x}
" />
$\frac{d}{dx} \arctan(u) = \frac{u'}{1+u^2}
$

in this case, $u = \sqrt{x}$