derivative question.

**Evan.Kimia** · February 19th 2010, 10:28 AM

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

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

**Evan.Kimia** · February 19th 2010, 10:29 AM

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.

**e^(i*pi)** · February 19th 2010, 10:33 AM

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

**ebaines** · February 19th 2010, 12:26 PM

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 $

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