- anonymous

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

- chestercat

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- anonymous

i have gotten down to: \[\sqrt{x}(6x ^{2})+(2x ^{3}-1)+(x ^{2})\]

- anonymous

just not sure how to simplify from here.

- cwrw238

sqrtx * 6x^2 + (1/2) x^(-1/2) * (2x^3 - 1)

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## More answers

- anonymous

√x(6x^2)+(2x^3−1)+(x^2) is what i got.

- cwrw238

derivative of sqrtx or x^(1/2) is (1/2) x^(-1/2)

- anonymous

the deriviate of \[\sqrt{x} = x ^{2}\]

- anonymous

i think? lol, that's what my prof wrote down.

- cwrw238

the rule is
if f(x) = ax^n , f'(x) = an x^(n-1)

- cwrw238

if n = 1/2, n-1 = -1/2

- anonymous

so it's \[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] ??

- cwrw238

yes

- anonymous

okay im still not sure how to further simplfy from there.

- anonymous

wait, i don't even need the deriviate of the square root of x ..

- anonymous

nevermind i do lol anyways,

- cwrw238

which can be written as 1 / 2x^(1/2)
sorry can't help anymore - gotta go out

- anonymous

okay.

- anonymous

so im left with
\[x ^{\frac{ 1 }{ 2 }}(6x ^{2})+(2x-1)(\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\]
i need help further simplifying.

- anonymous

\[\huge 6x^2\sqrt x + \frac{2x^3-1}{2\sqrt x}\]add them ~

- anonymous

why 2 root x?

- anonymous

because\[\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\]

- anonymous

\[(fg)'=f'g+g'f\] with
\[f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},g(x)=2x^3-1,g'(x)=6x^2\]

- anonymous

yes i got all that, im just not sure how to further simply once im at the last step with all deriviates completed.

- anonymous

i have everything over 2 root x

- anonymous

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

- anonymous

i know all that but what im having trouble with is simplyfying
\[\sqrt{x}(6x ^{2})+2x ^{3}-1 /2\sqrt{x}\]

- anonymous

\[\huge 6x^2 \sqrt x + \frac{2x^3-1}{2 \sqrt x}\] you got to here, now just make the denominators common and add them

- anonymous

i have \[(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}-1 all \over 2\sqrt{x}\]

- anonymous

i cant get any further than that.

- anonymous

\[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]

- anonymous

no idea

- anonymous

12x^2 root x ?

- anonymous

???

- anonymous

\[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]

- anonymous

why x cubed and not x squared?

- anonymous

because~\[\huge 6x^2 \times 2 = 12x^2\]\[\huge \sqrt x \times \sqrt x = x\]\[\huge 12x^2 \times x = 12 x^3\]
then
\[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]

- anonymous

thank-you! then i can just rationalize the denominator and ill be done.

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