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Alice_Dump

  • 2 years ago

Ppl of the world.. here comes the Next question -..-

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  1. Alice_Dump
    • 2 years ago
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    its about the definition of the derivative >>>

  2. SmithsAccount
    • 2 years ago
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    were all beaming with anticipation...

  3. Alice_Dump
    • 2 years ago
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    Apply the definition of the d derivative to compute \[f \prime (x) \] where \[f(x)=3x ^{2}+x-2\]

  4. rvgupta
    • 2 years ago
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    6x+1

  5. AravindG
    • 2 years ago
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    6x+1

  6. Alice_Dump
    • 2 years ago
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    i know .. \[f'(x)= 6x +1\] .. but my teacher said do it with details like long way to get this answer

  7. Alice_Dump
    • 2 years ago
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    -..- i don't know how

  8. AravindG
    • 2 years ago
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    i see well |dw:1349804778192:dw|

  9. Alice_Dump
    • 2 years ago
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    XD ur hand-writting is Funnie XP

  10. AravindG
    • 2 years ago
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    lim delta x tends to 0

  11. SmithsAccount
    • 2 years ago
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    ikr!

  12. Alice_Dump
    • 2 years ago
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    O_O sorry didn't mean to .. oops im sorry >> >_< sowwie

  13. amorfide
    • 2 years ago
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    you have a function of x F(X)=3x²+x−2 to differentiate you multiply the coefficient of the unknown by the power of the unknown and decrease the power of the unknown by 1 for example f(x)=x² f'(x)=2(1)x^2-1=2x so f(x)=x f'(x)=1(1)x^1-1=x^0=1 if you have f(x)=2x² f'(x)=2(2)x^2-1=4x does this help? |dw:1349804820461:dw|

  14. AravindG
    • 2 years ago
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    np @Alice_Dump did u get that ?

  15. Alice_Dump
    • 2 years ago
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    -..- ppl brb .. gemme 5 minutes i need to do something .

  16. Alice_Dump
    • 2 years ago
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    wait okie >_<

  17. UnkleRhaukus
    • 2 years ago
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    \[f'(x)=2\times 3x^{2-1} +1\times x^{1-1}-0\times2x^{0-1}=6x +1\]

  18. rvgupta
    • 2 years ago
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    2*3x^(2-1)+x(1-1)

  19. satellite73
    • 2 years ago
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    \[f(x)=3x^2+x-2\] \[f(x+h)=3(x+h)^2+(x+h)-2=3(x^2+2xh+h^2)+x+h-2\] \[=3x^2+6xh+3h^2+x+h-2\] \[f(x+h)-f(x)=3x^2+6xh+3h^2+x+h-2-(32x^2+x-2)\] \[=6xh+h+3h^2\]

  20. rvgupta
    • 2 years ago
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    2 being constant will disappear

  21. satellite73
    • 2 years ago
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    divide by \(h\) and get \[\frac{f(x+h)-f(x)}{h}=6x+1+3h\] take the limit as \(h\to 0\) get \(6x+1\)

  22. satellite73
    • 2 years ago
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    you needed it done by hand right? not power rule it is a bunch of algebra, that is all only last step is calc, and that is taking the limit by replacing \(h\) by \(0\)

  23. AravindG
    • 2 years ago
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    indeed @satellite73 has illustrated the given problem nicely with the formula i have given you ...i think that clears evrything

  24. satellite73
    • 2 years ago
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    question said "Apply the definition of the derivative to compute"

  25. satellite73
    • 2 years ago
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    @AravindG thanks

  26. AravindG
    • 2 years ago
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    yw :)

  27. Alice_Dump
    • 2 years ago
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    :} back ..

  28. Alice_Dump
    • 2 years ago
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    i got the question now :] thnx ppl ^_^'

  29. Alice_Dump
    • 2 years ago
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    heehehe

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