## MTALHAHASSAN2 one year ago Differentiate each of the following function: 2x^3-1/x^2

1. MTALHAHASSAN2

F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)

2. MTALHAHASSAN2

So the derivative of the given function be like that right

3. campbell_st

well I'd start by writing both terms using index notation $y = 2x^3 - x^{-2}$ now apply the rule for differentiation to both terms $if~~ y = x^n ~~~y' = nx^{n - 1}$

4. MTALHAHASSAN2

Oh ok but you miss the -1

5. MTALHAHASSAN2

2x^3-1

6. campbell_st

no I didn't I used index notation $\frac{1}{x^2} = x^{-2}$

7. campbell_st

so is the question 1. $y = 2x^3 - \frac{1}{x^2}$ or 2. $y = 2x^{3 - \frac{1}{x^2}}$

8. MTALHAHASSAN2

Both are wrong

9. MTALHAHASSAN2

X^2 also a denominater of 2x^3

10. MTALHAHASSAN2

@campbell_st

11. campbell_st

ok... so its $\frac{2x^3 -1}{x^2}$

12. MTALHAHASSAN2

Yes

13. MTALHAHASSAN2

F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)

14. campbell_st

ok so there are a couple of ways to do this... 1. use the quitient rule... or 2. split it into 2 fractions... $y = \frac{2x^3}{x^2} - \frac{1}{x^2}$ so simplifying and using index notation you get $y = 2x - x^{-2}$ so the derivative is $y' = 2 + \frac{2}{x^3}$

15. campbell_st

using the method you ahve attempted, which looks like the quotient rule you forgot the denominator if $y = \frac{f(x)}{g(x)}$ then $y' = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}$

16. MTALHAHASSAN2

No I guess that is product rule

17. MTALHAHASSAN2

Y′=f(x)1h(x)+f(x)h(x)1

18. MTALHAHASSAN2

@campbell_st

19. campbell_st

well if you are using the product rule you need to rewrite it using negative powers for the denominator $\frac{2x^3 -1}{x^2} = (2x^3 -1)(x^{-2})$ which will make a big difference

20. MTALHAHASSAN2

Oh ok but can I use any of them

21. MTALHAHASSAN2

I mean the rules

22. campbell_st

you can use any of the rules, product rule, but careful with the power in the denominator quotient rule... normal diffierentiation rule after simplifying... they will all give the same answer

23. MTALHAHASSAN2

Wait then what is ur answer

24. MTALHAHASSAN2

F1(x)= 3(2x^2-1)(x)^-2+(2x^3-1)-2x

25. MTALHAHASSAN2

F1(x)=6x^-2(x^2-1)-4x(x^3-1)

26. MTALHAHASSAN2

@campbell_st

27. anonymous

You're making it harder than it is

28. anonymous

$y=\frac{2x^3-1}{x^2} \iff y = (2x^3-1)(x^{-2})$

29. anonymous

$y'=2\cdot 3x^2 \cdot x^{-2} + (-2x^{-3})(2x^3-1)$$y'=2\cdot 3x^{2-2} -2x^{-3}(2x^3)+2x^{-3}$$y'=6x^{0}-(2\cdot 2)x^{0} +2x^{-3}$$y'=6-4+2x^{-3}$$y'= ... 30. anonymous \[y' =....$ **

31. triciaal

do you have any questions?

32. MTALHAHASSAN2

Can someone plz explain this question again to me??

33. triciaal

suggest you use parenthesis to make sure your question is clear

34. triciaal

|dw:1444369433856:dw|

35. MTALHAHASSAN2

I am having trouble after that positive sign.

36. triciaal

3 approaches with practice you can decide which is better for you. end with the same answer.

37. MTALHAHASSAN2

I will go with easier one

38. triciaal

1) the quotient rule 2)rewrite with index use product rule and chain rule 3) divide and do each term

39. MTALHAHASSAN2

Wait what you mean by the index

40. triciaal

that's the point they are all easy. with practice you will determine when to use which one or you can just stick to one format. personal preference.

41. anonymous

I used the product rule: $$f' \cdot g + g' \cdot f$$

42. MTALHAHASSAN2

Can you please explain the second one

43. triciaal

what zepdrix did and also campbell above it's rewriting to use the product rule

44. MTALHAHASSAN2

I know the product one

45. triciaal

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46. MTALHAHASSAN2

Got u

47. triciaal

good

48. MTALHAHASSAN2

So lets us the product on e

49. triciaal

|dw:1444370231542:dw|

50. MTALHAHASSAN2

Ok, looks good

51. triciaal

|dw:1444370312351:dw|

52. MTALHAHASSAN2

Ok

53. triciaal

so you just need to know how to apply the rules and practice

54. triciaal

additional note you can use substitution to apply the chain rule

55. triciaal

any final questions on this?

56. triciaal

ok goodnight