Mathematics OpenStudy (anonymous):

what is f'(x) of the (sqrt) (5x) OpenStudy (anonymous):

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

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

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

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

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

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