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mustry

  • 2 years ago

any one help me abt calculus

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  1. Nimi404
    • 2 years ago
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    Wat do mean

  2. Nimi404
    • 2 years ago
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    Calculate wat

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

  4. AravindG
    • 2 years ago
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    you need differential?

  5. mustry
    • 2 years ago
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    d(uv)dx=udvdx+vdudx

  6. AravindG
    • 2 years ago
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    if you need to differentiate use product rule

  7. Nimi404
    • 2 years ago
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    Hey Aravin you again

  8. AravindG
    • 2 years ago
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    yep apply that

  9. AravindG
    • 2 years ago
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    @Nimi404 calculus is higher mathematics it doesnt mean calculate :P

  10. mustry
    • 2 years ago
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    how do derive this formula

  11. AravindG
    • 2 years ago
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    you mean you need to derive product rule ?

  12. Nimi404
    • 2 years ago
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    Ooh then

  13. mustry
    • 2 years ago
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    Find the derivative of y = (x3 − 6x)(2 − 4x3)

  14. AravindG
    • 2 years ago
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    confused ...can u be clear ?

  15. mustry
    • 2 years ago
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    please help me this

  16. mustry
    • 2 years ago
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    Find the derivative of y = (x3 − 6x)(2 − 4x3)

  17. Yahoo!
    • 2 years ago
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    u know Product Rule ? @mustry

  18. mustry
    • 2 years ago
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    yap

  19. AravindG
    • 2 years ago
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    |dw:1356793984995:dw|

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

  21. mustry
    • 2 years ago
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    then

  22. AravindG
    • 2 years ago
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    then used product rule

  23. mustry
    • 2 years ago
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    thankx but can u derive it product rule?

  24. Callisto
    • 2 years ago
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    Derive the product rule?

  25. mustry
    • 2 years ago
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    yes

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

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