derivative of square root

[Guest]

the problem: find f'(1) when f(x) is the squareroot of (x + squareroot of x)

i divided it up into: squareroot of x + squareroot of squareroot of x

so the derivative of the: squareroot of x is 1/2(squareroot of x) f'(1) = 1/2

now for the derivative of the: squareroot of squareroot of x i used the chain rule to get the: squareroot of 1/2x^-.5 and that is as far as i could get.

 

[pka]

\(\displaystyle \L
f(x) = \sqrt {x + \sqrt x } \quad \Rightarrow \quad f'(x) = \frac{1}{{2\sqrt {x + \sqrt x } }}\left[ {1 + \frac{1}{{2\sqrt x }}} \right]\)

 

[Guest]

ahh i forgot you have to multiply by the derivative of whats inside the square root.

thank you very much !