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mustry Group Title

any one help me abt calculus

  • one year ago
  • one year ago

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  1. Nimi404 Group Title
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    Wat do mean

    • one year ago
  2. Nimi404 Group Title
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    Calculate wat

    • one year ago
  3. mustry Group Title
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    If we have a product like y = (2x2 + 6x)(2x3 + 5x2)

    • one year ago
  4. AravindG Group Title
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    you need differential?

    • one year ago
  5. mustry Group Title
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    d(uv)dx=udvdx+vdudx

    • one year ago
  6. AravindG Group Title
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    if you need to differentiate use product rule

    • one year ago
  7. Nimi404 Group Title
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    Hey Aravin you again

    • one year ago
  8. AravindG Group Title
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    yep apply that

    • one year ago
  9. AravindG Group Title
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    @Nimi404 calculus is higher mathematics it doesnt mean calculate :P

    • one year ago
  10. mustry Group Title
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    how do derive this formula

    • one year ago
  11. AravindG Group Title
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    you mean you need to derive product rule ?

    • one year ago
  12. Nimi404 Group Title
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    Ooh then

    • one year ago
  13. mustry Group Title
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    Find the derivative of y = (x3 − 6x)(2 − 4x3)

    • one year ago
  14. AravindG Group Title
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    confused ...can u be clear ?

    • one year ago
  15. mustry Group Title
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    please help me this

    • one year ago
  16. mustry Group Title
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    Find the derivative of y = (x3 − 6x)(2 − 4x3)

    • one year ago
  17. Yahoo! Group Title
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    u know Product Rule ? @mustry

    • one year ago
  18. mustry Group Title
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    yap

    • one year ago
  19. AravindG Group Title
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    |dw:1356793984995:dw|

    • one year ago
  20. AravindG Group Title
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    where i took x^3-6x as first function and 2-4x^3 as second function

    • one year ago
  21. mustry Group Title
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    then

    • one year ago
  22. AravindG Group Title
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    then used product rule

    • one year ago
  23. mustry Group Title
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    thankx but can u derive it product rule?

    • one year ago
  24. Callisto Group Title
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    Derive the product rule?

    • one year ago
  25. mustry Group Title
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    yes

    • one year ago
  26. Callisto Group Title
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    Okay...*Take a deep breath*\[\frac{d}{dx}(uv)\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)\color{red}{+u(x+h)v(x)-u(x+h)v(x)}-u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]-v(x)[u(x+h)-u(x)]}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]}{h}-\lim_{h \rightarrow 0}\frac{v(x)[u(x+h)-u(x)]}{h}\]\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)+v(x)}{h}-v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}\]\[=u(x)v'(x)+u'(x)v(x)\]

    • one year ago
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