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If f(x)=the square root of 6x, find f'(x).
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\[f(x) = (6x)^{1 \over 2} \]Hint: use power rule to simplify.
if f(x)=6x^1/2 then the derivitive is simply 3x^-1/2
\[ {d \over dx}x^n = nx^{n - 1}\]
fprime=3/\[\sqrt{x}\]
I tried 3x^(-1/2) already but it says it's wrong. I also need to find f'(6).
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actually...it's not 3(x)^-1/2 \[\frac{d}{dx} \sqrt{g(x)} \implies \frac {g'(x)}{2\sqrt{g(x)}}\] so in this case..g(x) is 6x and g'(x) is 6 so you have \[\implies \frac{6}{2\sqrt{6x}}\] simplify to get the desired answer
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