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\[\sqrt{x}(2x ^{3}-1)\] Find the derivative using the product rule.

Mathematics
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i have gotten down to: \[\sqrt{x}(6x ^{2})+(2x ^{3}-1)+(x ^{2})\]
just not sure how to simplify from here.
sqrtx * 6x^2 + (1/2) x^(-1/2) * (2x^3 - 1)

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Other answers:

√x(6x^2)+(2x^3−1)+(x^2) is what i got.
derivative of sqrtx or x^(1/2) is (1/2) x^(-1/2)
the deriviate of \[\sqrt{x} = x ^{2}\]
i think? lol, that's what my prof wrote down.
the rule is if f(x) = ax^n , f'(x) = an x^(n-1)
if n = 1/2, n-1 = -1/2
so it's \[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] ??
yes
okay im still not sure how to further simplfy from there.
wait, i don't even need the deriviate of the square root of x ..
nevermind i do lol anyways,
which can be written as 1 / 2x^(1/2) sorry can't help anymore - gotta go out
okay.
so im left with \[x ^{\frac{ 1 }{ 2 }}(6x ^{2})+(2x-1)(\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] i need help further simplifying.
\[\huge 6x^2\sqrt x + \frac{2x^3-1}{2\sqrt x}\]add them ~
why 2 root x?
because\[\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\]
\[(fg)'=f'g+g'f\] with \[f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},g(x)=2x^3-1,g'(x)=6x^2\]
yes i got all that, im just not sure how to further simply once im at the last step with all deriviates completed.
i have everything over 2 root x
plug and play (algebra) not to butt in, but \(\sqrt{x}\) is a very very common function, plus math teachers like to put it on tests and quizzes, so rather than use the power rule each time it would help your cause out to memorize \[\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\] like you know \(8\times 9=72\) for example it will save you time on exams if nothing else
i know all that but what im having trouble with is simplyfying \[\sqrt{x}(6x ^{2})+2x ^{3}-1 /2\sqrt{x}\]
\[\huge 6x^2 \sqrt x + \frac{2x^3-1}{2 \sqrt x}\] you got to here, now just make the denominators common and add them
i have \[(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}-1 all \over 2\sqrt{x}\]
i cant get any further than that.
\[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]
no idea
12x^2 root x ?
???
\[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]
why x cubed and not x squared?
because~\[\huge 6x^2 \times 2 = 12x^2\]\[\huge \sqrt x \times \sqrt x = x\]\[\huge 12x^2 \times x = 12 x^3\] then \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]
thank-you! then i can just rationalize the denominator and ill be done.

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