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HELLP!!! What's the derivative (F'(x) ) of F(x) (√x-3+x^2)/4x ??
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\[f(x)=\frac{ \sqrt{x^2+x-3} }{ 4x }\] is this right? your equation is a little vague...
no the radical is just for the second x
\[\frac{\sqrt{x}-3+x^2}{4x}\]?
or is the \(-3\) under the radical as well? it makes a huge difference in the method
I think the -3 is under the radical as well
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guess i'll never know
Yes like you wrote it! :)
split it up into \[\frac{ \sqrt{x} }{ 4x } - \frac{3}{4x} + .25x\]
ok thanks :)
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