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

f'(x)

------

2sqrt[f(x)] ?

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- Feb 19th 2010, 10:28 AMEvan.Kimiaderivative question.
if f in sqrt(f(x)) is differentiable, would it equal to

f'(x)

------

2sqrt[f(x)] ? - Feb 19th 2010, 10:29 AMEvan.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. - Feb 19th 2010, 10:33 AMe^(i*pi)
- Feb 19th 2010, 12:26 PMebaines
Not quite right. You should have:

$\displaystyle

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

$

$\displaystyle

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

$

An easier way is to realize that

$\displaystyle \sqrt {x^3} = x^{3/2}

$

So you get right to:

$\displaystyle

f'(x) = \frac 3 2 x^{1/2} = \frac 3 2 \sqrt x

$