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bmelyk
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\[\sqrt{x}(2x ^{3}1)\]
Find the derivative using the product rule.
 2 years ago
 2 years ago
bmelyk Group Title
\[\sqrt{x}(2x ^{3}1)\] Find the derivative using the product rule.
 2 years ago
 2 years ago

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

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
just not sure how to simplify from here.
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
sqrtx * 6x^2 + (1/2) x^(1/2) * (2x^3  1)
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
√x(6x^2)+(2x^3−1)+(x^2) is what i got.
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
derivative of sqrtx or x^(1/2) is (1/2) x^(1/2)
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
the deriviate of \[\sqrt{x} = x ^{2}\]
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
i think? lol, that's what my prof wrote down.
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
the rule is if f(x) = ax^n , f'(x) = an x^(n1)
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
if n = 1/2, n1 = 1/2
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
so it's \[\frac{ 1 }{ 2 }x ^{\frac{ 1 }{ 2 }}\] ??
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
okay im still not sure how to further simplfy from there.
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
wait, i don't even need the deriviate of the square root of x ..
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
nevermind i do lol anyways,
 2 years ago

cwrw238 Group TitleBest 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
 2 years ago

bmelyk Group TitleBest 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.
 2 years ago

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.1
\[\huge 6x^2\sqrt x + \frac{2x^31}{2\sqrt x}\]add them ~
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
why 2 root x?
 2 years ago

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.1
because\[\frac{1}{2}x^{1/2} = \frac{1}{2\sqrt x}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[(fg)'=f'g+g'f\] with \[f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},g(x)=2x^31,g'(x)=6x^2\]
 2 years ago

bmelyk Group TitleBest 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.
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
i have everything over 2 root x
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

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

nphuongsun93 Group TitleBest 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
 2 years ago

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

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
i cant get any further than that.
 2 years ago

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.1
\[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
12x^2 root x ?
 2 years ago

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
why x cubed and not x squared?
 2 years ago

nphuongsun93 Group TitleBest 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 \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
thankyou! then i can just rationalize the denominator and ill be done.
 2 years ago
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