amy0799 one year ago if g(x)=x^2f'(x)

1. amy0799

where $f(x)=\frac{ x-cosx }{ x }$ find the average rate of change of g on [pi, 2pi]

2. SolomonZelman

differentiate the f(x).

3. SolomonZelman

$$\large\color{black}{ \displaystyle f(x) =\frac{x-\cos x}{x} }$$ $$\large\color{black}{ \displaystyle f(x) =1-\frac{\cos x}{x} }$$ then derivative of 1 is 0, and derivative of the second peace will go by quotient rule (or you can use product or logarithmic)

4. SolomonZelman

$$\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }$$

5. SolomonZelman

6. amy0799

why did you take the 2nd derivative?

7. SolomonZelman

I didn't take the second derivative.

8. amy0799

oh wait I misread ur sentence, sorry

9. SolomonZelman

My function is: $$\large\color{black}{ \displaystyle f(x) =\frac{x-\cos x}{x} }$$ then I simplify it algebraically, to get: $$\large\color{black}{ \displaystyle f(x) =1-\frac{\cos x}{x} }$$ And now, We have to differentiate it (once). $$\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }$$

10. SolomonZelman

if it is not making sense, then you are always welcome to express your concerns regarding the problem.

11. amy0799

so then $g(x)=x ^{2}(\frac{ xsinx+cosx}{ x ^{2} })$

12. amy0799

@SolomonZelman

13. SolomonZelman

yes, very good, and you can tell that x² will can cancel

14. SolomonZelman

Then the remaining part is an algebraic task of fining the slope (of the secant) between $$($$ 2π,f(2π) $$)$$ and $$($$ π,f(π) $$)$$

15. amy0799

I put 2pi and pi into xsinx+cosx?

16. amy0799

@SolomonZelman

17. SolomonZelman

Yes, that is what you do:)

18. amy0799

(2pi,1) (pi,-1) is that correct?

19. SolomonZelman

Yes

20. SolomonZelman

then the slope formula...

21. amy0799

22. SolomonZelman

|dw:1442437619280:dw|

23. SolomonZelman

Not the other way around

24. amy0799

oh oops so 2/pi?

25. SolomonZelman

Yup

26. amy0799

do u mind helping me with another problem?

27. SolomonZelman

Yes, if I will be able to.

28. amy0799

find $\lim_{h \rightarrow 0} \frac{ \csc(\frac{ \pi }{ 6 }+h)-\csc\frac{ \pi }{ 6 }}{ h }$ @SolomonZelman

29. SolomonZelman

Oh boy, first principles of differentiation.... (you will get to the point, when this will be only useful to perhaps prove derivtive of e^x, but I guess right now we can't apply the easy properties right... will go with that limit definition... i have to refresh my page. sorry)

30. amy0799

when I did this I got 2 and I'm not sure if that's correct

31. SolomonZelman

Something was going on with my PC, it is kind of off.

32. SolomonZelman

Ok, I think I am back.

33. SolomonZelman

With $$f(x)=\csc(x)$$ $$\large\color{slate}{\displaystyle f'(x)=\lim_{h \rightarrow ~0}\frac{\csc(x+h)-\csc(x)}{h}}$$ $$\large\color{slate}{\displaystyle f'(\pi/6)=\lim_{h \rightarrow ~0}\frac{\csc(\pi/6+h)-\csc(\pi/6)}{h}}$$

34. SolomonZelman

You don't get 2 for that. I really don't think so.

35. SolomonZelman

I get -2√3

36. amy0799

how?

37. SolomonZelman

I applied the chain rule: $$\displaystyle \frac{d}{dx} (\csc x)=\frac{d}{dx} \left((\sin x)^{-1}\right)=(-1)\sin^{-2}x \times \color{blue}{\cos x} \\=-\csc^2x \cos x=-\cot x\csc x$$

38. SolomonZelman

Then I plugged π/6 instead of x.

39. SolomonZelman

This is just for check...

40. SolomonZelman

But, I guess you have to use that lim h-> 0 rule, right?

41. amy0799

where did you get the cosx from?

42. SolomonZelman

have you studied the "Chain Rule" ?

43. SolomonZelman

If not, then don't worry about my application of the chain rule...

44. amy0799

yes

45. SolomonZelman

ok, so you want to do the limit?

46. amy0799

oh wait I see how you got it.

47. SolomonZelman

$$\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(\frac{\pi}{6}+h)-\csc(\frac{\pi}{6})}{h}}$$

48. SolomonZelman

Still want to do the limit?

49. amy0799

sure. I don't understand how you got $-\frac{ 2 }{ \sqrt{3} }$

50. SolomonZelman

I didn't.

51. SolomonZelman

I said tha I got -2√3

52. SolomonZelman

We have to do the limit, don't we?

53. amy0799

I believe so

54. SolomonZelman

$$\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(x+h)-\csc(x)}{h}}$$ $$\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{1}{\sin(x+h)}-\dfrac{1}{\sin(x)}}{h}}$$ $$\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)-\sin(x+h) }{\sin(x+h)\sin(x)}}{h}}$$ $$\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)-\sin(x)\cos(h) -\sin(h)\cos(x)}{[\sin(x)\cos(h) +\sin(h)\cos(x)]\sin(x)}}{h}}$$

55. SolomonZelman

this is too much for me... going through all of this limit.

56. amy0799

I'm sorta getting lost on what ur doing. does finding the limit get u $-2\sqrt{3}?$

57. SolomonZelman

|dw:1442439909186:dw|

58. SolomonZelman

If, finding the derivative through the chain rule, and plugging π/6 gives -2√3, then the limit must give the same.

59. SolomonZelman

I drew two identities that you will need to get rid of h in the denominator.... It is just not right tho, if I will type all of this for you, according to the site's policy-;9

60. amy0799

oh then I guess we don't have to find the limit. can you explain how you −2√3 from the chain rule?

61. amy0799

how did you get −cotxcscx from csc^2xcosx?

62. SolomonZelman

Do you get the essence of the chain rule? (before we go on to our problem)

63. amy0799

yes

64. SolomonZelman

Ok, so we are differentiating $$\csc x$$ (with respect to x) we know that: $$\csc x = 1/\sin x = \left( \sin x \right)^{-1}$$ right?

65. amy0799

right

66. SolomonZelman

So, we are going to differentiate *THAT* now.

67. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] }$$

68. SolomonZelman

Just as, $$\large\color{black}{ \displaystyle \frac{d }{dx} \left[ x ^{-1}\right]=(-1)\cdot x^{-1-1}=-x^{-2} }$$ So is, $$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1}=-\left( \sin x \right)^{-2}}$$ However, there is a *BUT* to that.

69. SolomonZelman

We must multiply times the derivative of the inner function, times the derivative of sin x.

70. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}$$

71. SolomonZelman

this in blue, is the CHAIN RULE, that we must apply

72. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}$$ $$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \left( \sin x \right)^{-2} \color{blue}{\times \cos x}}$$

73. SolomonZelman

so far so good?

74. amy0799

yup

75. SolomonZelman

then, you just simplify it: $$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \left( \sin x \right)^{-2} \color{blue}{\times \cos x}}$$ $$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \csc^2 x \color{blue}{\times \cos x}}$$ and thus we know that: $$\large\color{black}{ \displaystyle \frac{d }{dx} \left[\csc x\right] =- \csc x \cot x}$$

76. SolomonZelman

then, you just plug in π/6, which should not be a problem.

77. amy0799

I don't understand how you got −csc^2x×cosx

78. SolomonZelman

well, sin$$^{-2}$$x is same as 1/sin²x, which is csc²x. And the one in front came from the power rule from the very beginning

79. SolomonZelman

And cos(x), as I explained, is the chain rule.

80. amy0799

oh ok. so I have 2 more problems I need help with, can u help me? if not that's ok.