## anonymous 5 years ago can some one help me diferentiate x^1/2+x^-4/3

1. anonymous

in respect to what?

2. anonymous

to y

3. anonymous

$y = x ^{1/2} + x ^{-4/3}$ is this the original function?

4. anonymous

no the original function is y=sqrt of x + 1/cube root of x^4

5. anonymous

write it as an equation because now im conufse

6. anonymous

$\sqrt{x}+1/\sqrt[3]{x^4}$

7. anonymous

dy/dx = 1/2 x^(-1/2) -4/3 x^-7/3

8. anonymous

rewrite it as $y= x ^{1/2} + 1 /(x ^{4/3})$ $y= x ^{1/2} + (x ^{-4/3})$ $dy/dx= (1/2)x ^{-1/2} + (-4/3)x ^{-7/3}$

9. anonymous

understand how to get it?

10. anonymous

thats the first answer i got but i dont uderstand how i got it

11. anonymous

i understand the first step but am lost at the second

12. anonymous

are we using both the product rule and quotient rule here?

13. anonymous

no, just differentiate each term separately

14. anonymous

its the chain rule for example when you find the derivative of $x^a$ the answer is $a * x ^{a-1}$

15. anonymous

@Romero, I dont think its the chain rule since this is not a composite function this question only requires you to do basic differentiation using the rule "y=(a)x^n" differentiates to "dy/dx= (a)(n)x^(n-1)

16. anonymous

sorry for not getting the word right but the concept is the same in finding the derivative.

17. anonymous

see i actually got what romero got but tried to the product rule after

18. anonymous

You dont need to use the product or quotient rule here. That's only when you have 2 variables multiplying to dividing.

19. anonymous

yeah i just saw that

20. anonymous

thank you so much man i really appreciate it trying to cram for exam am going to have in a couple of days and this really helps

21. anonymous

Sure thing man. I think the problem you're having is rewriting out square roots. Here this might help.$\sqrt[a]{x^b} = x ^{(b/a)}$ if you have no number for a then a =2 also $1/x^a = 1*x ^{-a}$ It was hard for me to see it at first but in time this will become second hand to you.

22. anonymous

do you think with couple of more problems i think i got how to do them but i dont know if they are right

23. anonymous

sure

24. anonymous

y=cos(tanx) have to deferentiate

25. anonymous

you use the chain rule so you should get y'= sin(tan x) * sec^2 x

26. anonymous

why do i use the product rule here

27. anonymous

no you dont use the product rule you only have one x within tan that happens to be inside cos

28. anonymous

actually its suppose to be y'= -sin(tan x) * sec^2 x

29. anonymous

With this problem you need to use the chain rule because its a composite function. The chain rule is, if y=f(u) $dy/dx = dy/du*du/dx$ Let u=tanx y=cos(u) differentiates to -sin(u) tanx differentiates to $\sec ^{2}x$ therefore dy/dx= -sin(tanx)*$\sec ^{2}x$

30. anonymous

when y= cos(a) y' = cos(a)' * a' in this case a = tan(x) so y'= - sin(tan(x)) * tan(x)' where tan(x)'= sec^2 (x) so y'= - sin (tanx) * sec^2 x

31. anonymous

oh ok i am getting it

32. anonymous

how about 3x-2/$\sqrt{2x+1}$

33. anonymous

Use the quotient rule.

34. anonymous

how do i diferentiate $\sqrt{2x+1}$

35. anonymous

y= a/b where y'= (a' * b - b' * a)/ b^2 in this case a = 3x -2 $b= \sqrt{2x + 1}$

36. anonymous

so a'=3 right

37. anonymous

yes a'=3, b'=1/2(2x+1)^(-1/2)

38. anonymous

so am getting 3/2(2x+1)^-1/2-(2x+1)^-1/2 over 2x+1 is that right

39. anonymous

I'm not too sure if that's right maybe im just confused with the typing but I got 3(2x+1)^(1/2) - (3x+2)(2x+1)^(-1/2) all over 2x+1

40. anonymous

I found this really good summary of differentiation formulaes, it covers up most basic differentiation topics. Hope you find it useful for your exam coming up soon. :)

41. anonymous

AWESOME THANKS MAN

42. anonymous

how did you get rid of the 1/2 after you derived $\sqrt{2x+1}$

43. anonymous

when you differentiate root 2x+1, you use the chain rule. so its $\sqrt{2x+1}, dy/dx= (1/2)(2)(2x+1)^{-1/2}$ so the 2 cancels out with the 1/2

44. anonymous

ok cool

45. anonymous

thnxs for the help man

46. anonymous

no problemo ;) btw, good luck for yur exam

47. anonymous

thns man am gonna need it