anonymous
  • anonymous
\[\sqrt{x}(2x ^{3}-1)\] Find the derivative using the product rule.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i have gotten down to: \[\sqrt{x}(6x ^{2})+(2x ^{3}-1)+(x ^{2})\]
anonymous
  • anonymous
just not sure how to simplify from here.
cwrw238
  • cwrw238
sqrtx * 6x^2 + (1/2) x^(-1/2) * (2x^3 - 1)

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More answers

anonymous
  • anonymous
√x(6x^2)+(2x^3−1)+(x^2) is what i got.
cwrw238
  • cwrw238
derivative of sqrtx or x^(1/2) is (1/2) x^(-1/2)
anonymous
  • anonymous
the deriviate of \[\sqrt{x} = x ^{2}\]
anonymous
  • anonymous
i think? lol, that's what my prof wrote down.
cwrw238
  • cwrw238
the rule is if f(x) = ax^n , f'(x) = an x^(n-1)
cwrw238
  • cwrw238
if n = 1/2, n-1 = -1/2
anonymous
  • anonymous
so it's \[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] ??
cwrw238
  • cwrw238
yes
anonymous
  • anonymous
okay im still not sure how to further simplfy from there.
anonymous
  • anonymous
wait, i don't even need the deriviate of the square root of x ..
anonymous
  • anonymous
nevermind i do lol anyways,
cwrw238
  • cwrw238
which can be written as 1 / 2x^(1/2) sorry can't help anymore - gotta go out
anonymous
  • anonymous
okay.
anonymous
  • 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
  • anonymous
\[\huge 6x^2\sqrt x + \frac{2x^3-1}{2\sqrt x}\]add them ~
anonymous
  • anonymous
why 2 root x?
anonymous
  • anonymous
because\[\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\]
anonymous
  • 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
  • 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
  • anonymous
i have everything over 2 root x
anonymous
  • 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
  • anonymous
i know all that but what im having trouble with is simplyfying \[\sqrt{x}(6x ^{2})+2x ^{3}-1 /2\sqrt{x}\]
anonymous
  • 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
  • anonymous
i have \[(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}-1 all \over 2\sqrt{x}\]
anonymous
  • anonymous
i cant get any further than that.
anonymous
  • anonymous
\[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]
anonymous
  • anonymous
no idea
anonymous
  • anonymous
12x^2 root x ?
anonymous
  • anonymous
???
anonymous
  • anonymous
\[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]
anonymous
  • anonymous
why x cubed and not x squared?
anonymous
  • 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
  • anonymous
thank-you! then i can just rationalize the denominator and ill be done.

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