## anonymous 5 years ago what is f'(x) of the (sqrt) (5x)

1. anonymous

5/2*$\sqrt{5}^(-1/2)$ = 5/(2sqrt(5))

2. anonymous

$f(x)=5^{1/2} \Rightarrow f'(x)=\frac{1}{2}*5^{-1/2}=\frac{1}{2\sqrt{5}}$

3. anonymous

$\prime(x)=\sqrt{5x} = 1/2*5^-1/2*5 = 5/2\sqrt{5}$

4. anonymous

Aagh, I'm my answer is off. 1st Step (Power Rule): $f(x)={(5x)}^{1/2} \Rightarrow f'(x)=\frac{1}{2}*\sqrt{5}*{x}^{-1/2}=\frac{\sqrt{5}}{2\sqrt{x}}$

5. anonymous

Typically we rationalize the denominator - multiply top and bottom by x^(1/2) $=\frac{\sqrt{5}}{2\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{5x}}{2x}$