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OpenStudy (anonymous):

what is f'(x) of the (sqrt) (5x)

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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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