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Rohangrr Group Title

Use the limit definition to compute the derivative, f'(x), \[f(x)=\frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \]

  • 2 years ago
  • 2 years ago

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  1. mathslover Group Title
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    oh k progression or in progress? MAN AT WORK :P

    • 2 years ago
  2. chemENGINEER Group Title
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    Answer on Progress

    • 2 years ago
  3. waterineyes Group Title
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    Oh wow what a question...!!

    • 2 years ago
  4. mathslover Group Title
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    rohangrr is typing a reply...

    • 2 years ago
  5. mathslover Group Title
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    oh k here we go

    • 2 years ago
  6. Rohangrr Group Title
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    Use the limit definition to compute the derivative, f'(x), \[f(x)=\frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \]

    • 2 years ago
  7. Mimi_x3 Group Title
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    derivative from the first principle..?

    • 2 years ago
  8. waterineyes Group Title
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    I am not typing..

    • 2 years ago
  9. mathslover Group Title
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    |dw:1345962336537:dw|

    • 2 years ago
  10. Mimi_x3 Group Title
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    \[\frac{dy}{dx} = \lim_{x \rightarrow 0} \frac{f(x+h)-f(x)}{h} \]

    • 2 years ago
  11. Mimi_x3 Group Title
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    mathslover; its from the first principle..

    • 2 years ago
  12. mathslover Group Title
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    oh sorry

    • 2 years ago
  13. Rohangrr Group Title
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    \[\huge f'\quad (x)\quad =\lim _{ x\rightarrow 0 } \frac { f(x+\triangle x)-f(x) }{ \triangle x } \] \[\huge \quad =\lim _{ x\rightarrow 0 } \frac { \frac { 1 }{ 2 } (x+\triangle x)-\frac { 3 }{ 5 } -\{ \frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \} }{ \triangle x } \] What will be after this

    • 2 years ago
  14. hartnn Group Title
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    first of all the limit would be h->0 !! then u cancel,x/2 and 3/5 in numerator.....

    • 2 years ago
  15. hartnn Group Title
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    i mean delta x ->0

    • 2 years ago
  16. Rohangrr Group Title
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    Okay then

    • 2 years ago
  17. hartnn Group Title
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    then u will simply get:|dw:1345962738189:dw|

    • 2 years ago
  18. mathslover Group Title
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    the same I got but my method was wrong :(

    • 2 years ago
  19. Rohangrr Group Title
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    \quad =\lim _{ \triangle x\rightarrow 0 } \frac { \frac { 1 }{ 2 } x+\frac { 1 }{ 2 } \triangle x-\frac { 3 }{ 5 } -\frac { 1 }{ 2 } x+\frac { 3 }{ 5 } \} }{ \triangle x }

    • 2 years ago
  20. Rohangrr Group Title
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    is this the final answer

    • 2 years ago
  21. Rohangrr Group Title
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    |dw:1345963107126:dw|

    • 2 years ago
  22. Rohangrr Group Title
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    Srously IM confused

    • 2 years ago
  23. hartnn Group Title
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    |dw:1345963124504:dw|

    • 2 years ago
  24. Rohangrr Group Title
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    |dw:1345963183321:dw|

    • 2 years ago
  25. Rohangrr Group Title
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    |dw:1345963244684:dw|

    • 2 years ago
  26. myininaya Group Title
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    Do you still have a question?

    • 2 years ago
  27. Rohangrr Group Title
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    Hi @myininaya I didnt observe u here

    • 2 years ago
  28. myininaya Group Title
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    Yes. I'm here.

    • 2 years ago
  29. Rohangrr Group Title
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    Thanks guys

    • 2 years ago
  30. Callisto Group Title
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    First principle: \[f'(x) =\lim_{Δ x \rightarrow 0}\frac{f(x+Δ x)-f(x)}{Δ x}\] When \(f(x) = \frac{1}{2}x - \frac{3}{5}\), \[f'(x) =\lim_{Δ x \rightarrow 0}\frac{[\frac{1}{2}(x+Δ x)-\frac{3}{5}]-(\frac{1}{2}x - \frac{3}{5})}{Δ x}\] \[=\lim_{Δ x \rightarrow 0}\frac{\frac{1}{2}x+\frac{1}{2}Δ x-\frac{3}{5}-\frac{1}{2}x + \frac{3}{5}}{Δ x}\] \[=\lim_{Δ x \rightarrow 0}\frac{\frac{1}{2}Δ x}{Δ x}\] \[=\lim_{Δ x \rightarrow 0}\frac{\frac{1}{2}\cancel{Δ x}}{\cancel{Δ x}}\] \[=\lim_{Δ x \rightarrow 0}\frac{1}{2}\] \[=...\] Wow! I'm tooooo late!

    • 2 years ago
  31. Rohangrr Group Title
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    But u r still alive

    • 2 years ago
  32. waterineyes Group Title
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    But I am still in confusion now..

    • 2 years ago
  33. waterineyes Group Title
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    To Whom should I give the medal ??

    • 2 years ago
  34. hartnn Group Title
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    what confusion u have?

    • 2 years ago
  35. waterineyes Group Title
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    No, it is now clear...

    • 2 years ago
  36. hartnn Group Title
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    ok..

    • 2 years ago
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