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roselin Group Title

find the derivative of u=5x+1/2 sqrt of x

  • 2 years ago
  • 2 years ago

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  1. Lexxiie123 Group Title
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    what is sqrt

    • 2 years ago
  2. Lexxiie123 Group Title
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    is it square root?

    • 2 years ago
  3. roselin Group Title
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    yes

    • 2 years ago
  4. mark_o. Group Title
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    derivative of u=5x+(1/2) sqrt of x sometimes its easier to use the power rule, and rewriting it as u=5x+(1/2 )( x)^1/2 use power rule of the derivative

    • 2 years ago
  5. mark_o. Group Title
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    d(kx)/dx=k.,,dx^n/dx= nx^n-1

    • 2 years ago
  6. roselin Group Title
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    the problem is actually: |dw:1351567010524:dw|

    • 2 years ago
  7. mark_o. Group Title
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    u=5x + 1/(2sqrtx)?

    • 2 years ago
  8. roselin Group Title
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    \[u= 5x+1\div 2\sqrt{x}\]

    • 2 years ago
  9. mark_o. Group Title
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    it will be the same idea by using the power rule u=5x + 1/(2sqrtx) u=5x + (1/2)(x)^-1/2 now find the derivative

    • 2 years ago
  10. roselin Group Title
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    what if i use the quotient rule?

    • 2 years ago
  11. mark_o. Group Title
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    u=5x + (1/2)(x)^-1/2 is it du/dt=5+(1/2)(-1/2)x^-1/2 - 2/2) du/dt=5+(1/2)(-1/2)x^-3/2 ) ?

    • 2 years ago
  12. roselin Group Title
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    its find the derivative of the function \[u= 5x+1 \div \sqrt{2}\]

    • 2 years ago
  13. mark_o. Group Title
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    ok the quotient rule says \[\frac{ d(\frac{ U }{ V }) }{ dx }=\frac{ V \frac{ du }{ dx }-U \frac{ dV }{ dx } }{ v ^{2} }\]

    • 2 years ago
  14. roselin Group Title
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    \[u= 5x+1\div 2\sqrt{x}\]

    • 2 years ago
  15. roselin Group Title
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    so i must use the power rule?

    • 2 years ago
  16. mark_o. Group Title
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    arrange them like \[u=5x +(\frac{ 1 }{ 2 })(\frac{ 1 }{ x ^{\frac{ 1 }{ 2 }} })\]

    • 2 years ago
  17. roselin Group Title
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    okay

    • 2 years ago
  18. mark_o. Group Title
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    what did you get ?

    • 2 years ago
  19. roselin Group Title
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    give me a minute

    • 2 years ago
  20. mark_o. Group Title
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    lets try to work only on 1/x^1/2

    • 2 years ago
  21. mark_o. Group Title
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    u=1, v=x^1/2

    • 2 years ago
  22. roselin Group Title
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    where did u get the v from?

    • 2 years ago
  23. mark_o. Group Title
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    if you are using the quotient rule Dx(U/V)

    • 2 years ago
  24. roselin Group Title
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    wait the quotient rule is |dw:1351568697070:dw|

    • 2 years ago
  25. roselin Group Title
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    that is the quotient rule right?

    • 2 years ago
  26. mark_o. Group Title
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    lets try to work only on 1/x^1/2 therefore Dx(U/V)=Dx( 1/x^1/2) so here u=1 and v=x^1/2

    • 2 years ago
  27. roselin Group Title
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    its 2

    • 2 years ago
  28. roselin Group Title
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    sorry,

    • 2 years ago
  29. mark_o. Group Title
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    yes on youR formula g=V..,f=u , its the same..you guys have taught f and g but variables such u,v, w are used also locally and internationally

    • 2 years ago
  30. roselin Group Title
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    okay,

    • 2 years ago
  31. mark_o. Group Title
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    ok try to use your formula and you will arrive with the same answer anyway

    • 2 years ago
  32. roselin Group Title
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    i am solving it here, give me a minute i will post what i have

    • 2 years ago
  33. mark_o. Group Title
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    ok take your time..lol.. :D

    • 2 years ago
  34. roselin Group Title
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    \[-5x+6/ 2\sqrt{x}\]

    • 2 years ago
  35. roselin Group Title
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    i got that answer

    • 2 years ago
  36. roselin Group Title
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    u thre?

    • 2 years ago
  37. mark_o. Group Title
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    hm try to study this one let u=1 and v= x^1/2 Dx(1/(x)^1/2) x^1/2(Dx(1) - 1 Dx(x^1/2) = --------------------------- ( x^1/2)^2 0 - (1/2)x^-1/2 = ---------------- x -(1/2) = --------- x x^1/2 -(1/2) =-------------- x^3/2 -1 = ---------- 2 x^3/2

    • 2 years ago
  38. mark_o. Group Title
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    now lets go back to the prob derivative of u=5x+(1/2)( x)^1/2 1 du/dx= 5 - ------------ 2( 2x^3/2) 1 = 5- ------------- 4x^3/2 w/c is the same as doing it w/ power rule

    • 2 years ago
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