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MTALHAHASSAN2

  • one year ago

Differentiate each function. Do not expand an expression before differentiating. a)f(x)=(2x+3)^4

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  1. MTALHAHASSAN2
    • one year ago
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    f(x)=4(2x+3)^3

  2. MTALHAHASSAN2
    • one year ago
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    is it be like that

  3. LynFran
    • one year ago
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    use chain rule

  4. anonymous
    • one year ago
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    |dw:1443580871976:dw|

  5. anonymous
    • one year ago
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    where u(x)=2x+3 and n=4

  6. MTALHAHASSAN2
    • one year ago
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    oh ok f(x)= 4(2x+3)^3 (2) = 8(2x+3)^3

  7. anonymous
    • one year ago
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    good work

  8. MTALHAHASSAN2
    • one year ago
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    thnx

  9. MTALHAHASSAN2
    • one year ago
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    g(x)=(x^2-4)^3 =3(x^2-4)^2 =6x(x^2-4)^2 For this question Is it be like this!!

  10. dan815
    • one year ago
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    ayeee what tower is that in the back, lol why this place look so familiar!

  11. MTALHAHASSAN2
    • one year ago
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    lol its nagara falls!!!

  12. dan815
    • one year ago
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    hahahaha

  13. dan815
    • one year ago
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    thats skylone right

  14. anonymous
    • one year ago
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    nah clearly that looks like the sydney harbour bridge

  15. dan815
    • one year ago
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    lmao pls xD

  16. anonymous
    • one year ago
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    yes you are right @MTALHAHASSAN2 with the other question

  17. MTALHAHASSAN2
    • one year ago
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    ok thnx again

  18. Zarkon
    • one year ago
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    you should not write g(x)=(x^2-4)^3 =3(x^2-4)^2 =6x(x^2-4)^2 as none of those are equal to each other

  19. MTALHAHASSAN2
    • one year ago
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    how about this one: f(x)= (pie^2-x^2)3

  20. MTALHAHASSAN2
    • one year ago
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    3(pie^2-x^2)2

  21. MTALHAHASSAN2
    • one year ago
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    3(pie^2-x^2)^2 (2pie-x)

  22. zepdrix
    • one year ago
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    \[\large\rm f(x)= (\pi^2-x^2)^3\]Your first step looks good, applying power rule.\[\large\rm f'(x)= 3(\pi^2-x^2)^2\color{royalblue}{(\pi^2-x^2)'}\]But you also need chain rule.

  23. zepdrix
    • one year ago
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    Ah very close!

  24. zepdrix
    • one year ago
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    Derivative of x^2 is not x. And derivative of pi^2 is not 2pi.

  25. MTALHAHASSAN2
    • one year ago
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    i know its 2x

  26. zepdrix
    • one year ago
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    2pi is just a number. It's a fancy looking constant. Remember how to differentiate a constant? \(\large\rm \frac{d}{dx}3=?\)

  27. zepdrix
    • one year ago
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    pi^2 is just a number* blah typo :)

  28. MTALHAHASSAN2
    • one year ago
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    then is it be 6(pie^2-x^2)^2

  29. MTALHAHASSAN2
    • one year ago
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    @zepdrix is it right??

  30. MTALHAHASSAN2
    • one year ago
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    but at the back of the book they have -6 instead of 6

  31. zepdrix
    • one year ago
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    oh sorry i ran off for a sec >.<

  32. MTALHAHASSAN2
    • one year ago
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    oh ok np

  33. zepdrix
    • one year ago
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    \[\large\rm f'(x)= 3(\pi^2-x^2)^2\color{royalblue}{(\pi^2-x^2)'}\]\[\large\rm f'(x)= 3(\pi^2-x^2)^2\color{orangered}{(0-2x)}\]Ya, don't forget about the negative in front of the x^2

  34. zepdrix
    • one year ago
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    \[\large\rm f'(x)= 3(\pi^2-x^2)^2(-2x)\]\[\large\rm f'(x)= -6x(\pi^2-x^2)^2\]Understand why there is a negative? :o

  35. MTALHAHASSAN2
    • one year ago
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    yeah got you and thnx a lot

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