1. ## derivative question.

if f in sqrt(f(x)) is differentiable, would it equal to

f'(x)
------
2sqrt[f(x)] ?

2. trying this out with x^3, i get 2(x^3)-1*(3x^2), and i think this would yield....

3x^2
------
2(x^3)

I know this is a dumb simplification problem but i just wanted some clarification. thanks.

3. Originally Posted by Evan.Kimia
if f in sqrt(f(x)) is differentiable, would it equal to

f'(x)
------
2sqrt[f(x)] ?
Let $y = \sqrt{f(x)}$

$y^2 = f(x)$

$2y\frac{dy}{dx} = f'(x)$

$\frac{dy}{dx} = \frac{f'(x)}{2y} = \frac{f'(x)}{2\sqrt{f(x)}}$

I get the same answer as you but I'm not 100% confident it's right

4. Originally Posted by Evan.Kimia
trying this out with x^3, i get 2(x^3)-1*(3x^2), and i think this would yield....

3x^2
------
2(x^3)

I know this is a dumb simplification problem but i just wanted some clarification. thanks.
Not quite right. You should have:

$
f(x) = \sqrt {x^3}
$

$
f'(x) = \frac {3x^2} {2 \sqrt{x^3}} = \frac 3 2 \sqrt x
$

An easier way is to realize that
$\sqrt {x^3} = x^{3/2}
$

So you get right to:
$
f'(x) = \frac 3 2 x^{1/2} = \frac 3 2 \sqrt x
$