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

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

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

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

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

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

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

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

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

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

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

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

bmelyk
 3 years ago
Best ResponseYou've already chosen the best response.0nevermind i do lol anyways,

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.0which can be written as 1 / 2x^(1/2) sorry can't help anymore  gotta go out

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

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

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

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

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

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0plug 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

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

nphuongsun93
 3 years ago
Best 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

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

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

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

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

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

nphuongsun93
 3 years ago
Best ResponseYou've already chosen the best response.1because~\[\huge 6x^2 \times 2 = 12x^2\]\[\huge \sqrt x \times \sqrt x = x\]\[\huge 12x^2 \times x = 12 x^3\] then \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]

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