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

1. bmelyk

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

2. bmelyk

just not sure how to simplify from here.

3. cwrw238

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

4. bmelyk

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

5. cwrw238

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

6. bmelyk

the deriviate of $\sqrt{x} = x ^{2}$

7. bmelyk

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

8. cwrw238

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

9. cwrw238

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

10. bmelyk

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

11. cwrw238

yes

12. bmelyk

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

13. bmelyk

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

14. bmelyk

nevermind i do lol anyways,

15. cwrw238

which can be written as 1 / 2x^(1/2) sorry can't help anymore - gotta go out

16. bmelyk

okay.

17. bmelyk

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

18. nphuongsun93

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

19. bmelyk

why 2 root x?

20. nphuongsun93

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

21. satellite73

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

22. bmelyk

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

23. bmelyk

i have everything over 2 root x

24. satellite73

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

25. bmelyk

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

26. nphuongsun93

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

27. bmelyk

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

28. bmelyk

i cant get any further than that.

29. nphuongsun93

$\huge 6x^2 \sqrt x \times 2 \sqrt x = ?$

30. bmelyk

no idea

31. bmelyk

12x^2 root x ?

32. bmelyk

???

33. nphuongsun93

$\huge \sqrt x \times \sqrt x = x$$\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3$

34. bmelyk

why x cubed and not x squared?

35. nphuongsun93

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

36. bmelyk

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