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

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bmelykBest ResponseYou've already chosen the best response.0
i have gotten down to: \[\sqrt{x}(6x ^{2})+(2x ^{3}1)+(x ^{2})\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
just not sure how to simplify from here.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
sqrtx * 6x^2 + (1/2) x^(1/2) * (2x^3  1)
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
√x(6x^2)+(2x^3−1)+(x^2) is what i got.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
derivative of sqrtx or x^(1/2) is (1/2) x^(1/2)
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
the deriviate of \[\sqrt{x} = x ^{2}\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i think? lol, that's what my prof wrote down.
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
the rule is if f(x) = ax^n , f'(x) = an x^(n1)
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
if n = 1/2, n1 = 1/2
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so it's \[\frac{ 1 }{ 2 }x ^{\frac{ 1 }{ 2 }}\] ??
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
okay im still not sure how to further simplfy from there.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
wait, i don't even need the deriviate of the square root of x ..
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
nevermind i do lol anyways,
 one year ago

cwrw238Best ResponseYou've already chosen the best response.0
which can be written as 1 / 2x^(1/2) sorry can't help anymore  gotta go out
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so im left with \[x ^{\frac{ 1 }{ 2 }}(6x ^{2})+(2x1)(\frac{ 1 }{ 2 }x ^{\frac{ 1 }{ 2 }}\] i need help further simplifying.
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge 6x^2\sqrt x + \frac{2x^31}{2\sqrt x}\]add them ~
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
because\[\frac{1}{2}x^{1/2} = \frac{1}{2\sqrt x}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
\[(fg)'=f'g+g'f\] with \[f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},g(x)=2x^31,g'(x)=6x^2\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
yes i got all that, im just not sure how to further simply once im at the last step with all deriviates completed.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i have everything over 2 root x
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i know all that but what im having trouble with is simplyfying \[\sqrt{x}(6x ^{2})+2x ^{3}1 /2\sqrt{x}\]
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge 6x^2 \sqrt x + \frac{2x^31}{2 \sqrt x}\] you got to here, now just make the denominators common and add them
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i have \[(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}1 all \over 2\sqrt{x}\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i cant get any further than that.
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
\[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
why x cubed and not x squared?
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.1
because~\[\huge 6x^2 \times 2 = 12x^2\]\[\huge \sqrt x \times \sqrt x = x\]\[\huge 12x^2 \times x = 12 x^3\] then \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
thankyou! then i can just rationalize the denominator and ill be done.
 one year ago
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