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Matthew071

  • one year ago

Given that f(x)=-2h(x)+(2x/h(x)); h(4)=4, h'(4)=-2. Find f'(4)

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  1. Matthew071
    • one year ago
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    well their is: a)13/2 b)7/2 c)-5/2 d)11/2 e)17/2

  2. Matthew071
    • one year ago
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    I tried solving for the derivative of f(x) but I must be doing something wrong or the awnser is not there

  3. myininaya
    • one year ago
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    So what did you get for f'?

  4. myininaya
    • one year ago
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    I will check it.

  5. Matthew071
    • one year ago
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    \[f'(x) = h(x) + -2h'(x)+ (\frac{ 2h(x)-(2x)h(x) }{ h(x)^2 }\]

  6. myininaya
    • one year ago
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    Also the answer is there. h(x) shouldn't be there. you are missing h' in your quotient

  7. myininaya
    • one year ago
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    (-2h)'=-2(h)'=-2h'

  8. Matthew071
    • one year ago
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    oh ok

  9. myininaya
    • one year ago
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    (2x/h)'=[(2x)'h-2x(h)']/h^2

  10. Matthew071
    • one year ago
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    so the awnser should be 19/2

  11. myininaya
    • one year ago
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    hmm... that isn't what I got...let me check my result.

  12. myininaya
    • one year ago
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    Oh you are still assuming h is there aren't you?

  13. myininaya
    • one year ago
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    (-2h)'=-2(h)'=-2h' not h-2h'

  14. Matthew071
    • one year ago
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    So how does it apply

  15. myininaya
    • one year ago
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    what do you mean?

  16. Matthew071
    • one year ago
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    well I am plugging in everything where \[h(4)+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = 4 +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)\]

  17. Matthew071
    • one year ago
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    correct?

  18. myininaya
    • one year ago
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    Well see I keep telling you (-2h)'=-2h' not h-2h'

  19. myininaya
    • one year ago
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    That first term you are putting in shouldn't be there.

  20. Matthew071
    • one year ago
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    Could you show me what you plugged in so I understand

  21. myininaya
    • one year ago
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    So you don't understand why (-2h)' is -2h' and not h-2h'?

  22. Matthew071
    • one year ago
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    No

  23. myininaya
    • one year ago
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    I'm liking I'm using the constant multiple rule.

  24. myininaya
    • one year ago
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    (cf)'=cf' Let k(x)=cf(x) To find k'(x) we use the formal definition of derivative: \[k'(x)=\lim_{\Delta x \rightarrow 0} \frac{k(x+\Delta x)-k(x)}{\Delta x}=\lim_{\Delta x \rightarrow 0} \frac{cf(x+\Delta x)-cf(x)}{\Delta x}\] Note: By the distributive property ab+ac=a(b+c) \[=\lim_{\Delta x \rightarrow 0} c \frac{f(x+\Delta x)-f(x)}{\Delta x} \] Note: By limit properties, we can do lim c g(x) = c lim g(x) \[=c \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \] Note: By definition of derivative that above is cf'(x).

  25. myininaya
    • one year ago
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    The constant multiple rule: k=cf k'=cf'

  26. myininaya
    • one year ago
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    So say k=ch then k'=ch' That c means constant The constant in front of h in your problem is -2 so k=-2h then k'=-2h'

  27. Matthew071
    • one year ago
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    So when I plug in for -2h' when h(4)=4 and h'(4)=-2

  28. myininaya
    • one year ago
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    Yep

  29. myininaya
    • one year ago
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    -2h'(x) x is 4. -2h'(4) -2h'(4)=-2(-2)=4 Now simplify your quotient and add.

  30. Matthew071
    • one year ago
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    Does this mean -2h'= -2(-2) or -2(4)

  31. Matthew071
    • one year ago
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    ok

  32. myininaya
    • one year ago
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    h' is h'(x)

  33. myininaya
    • one year ago
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    [-2h(x)] ' =-2 [ h(x)] ' =-2 h'(x) This has been what I have been saying but you wrote it as h(x)-2h'(x)

  34. myininaya
    • one year ago
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    \[\cancel{h(4)}+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = \cancel{4} +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)\]

  35. myininaya
    • one year ago
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    This is what you wrote earlier I'm telling you (-2h)'=-2h' not h-2h' do you understand now?

  36. Matthew071
    • one year ago
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    Oh ok Now I get what you mean by this

  37. myininaya
    • one year ago
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    I put the marks through the bad parts.

  38. myininaya
    • one year ago
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    so you understand the constant multiple rule?

  39. Matthew071
    • one year ago
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    Yes I do now understand the multiple rule and the awnser is 11/2

  40. myininaya
    • one year ago
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    That is right.

  41. Matthew071
    • one year ago
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    I just didnt understand it conceptually but now that I see it plugged in and used I do now

  42. myininaya
    • one year ago
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    Ok. Neatness.

  43. Matthew071
    • one year ago
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    Thank you so much

  44. myininaya
    • one year ago
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    Np.

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