Question

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

Answer 1

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

Answer 2

just not sure how to simplify from here.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

the rule is

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

Answer 9

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

Answer 10

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

Answer 11

yes

Answer 12

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

Answer 13

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

Answer 14

nevermind i do lol anyways,

Answer 15

which can be written as 1 / 2x^(1/2)

sorry can't help anymore - gotta go out

Answer 16

okay.

Answer 17

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

Answer 18

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

Answer 19

why 2 root x?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

i have everything over 2 root x

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

i cant get any further than that.

Answer 29

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

Answer 30

no idea

Answer 31

12x^2 root x ?

Answer 32

???

Answer 33

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

Answer 34

why x cubed and not x squared?

Answer 35

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

Answer 36

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