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needhelp07

  • 3 years ago

Problem : Find the derivatives of the function using the limit of definition y=x^4

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  1. irkiz
    • 3 years ago
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    still the same type of question? try going http://www.math.hmc.edu/calculus/tutorials/limit_definition/ they show you examples. just apply your question to how they solve it.

  2. irkiz
    • 3 years ago
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    It cant be that you were taught 2 other similar questions but still cant solve it.

  3. needhelp07
    • 3 years ago
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    im confused about the exponent 4

  4. mark_o.
    • 3 years ago
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    do you know how to do (a+b)^4?

  5. needhelp07
    • 3 years ago
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    no

  6. mark_o.
    • 3 years ago
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    its the binomial theorem or pascal triangle

  7. mark_o.
    • 3 years ago
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    1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 etc....

  8. mark_o.
    • 3 years ago
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    for example just look at the pascal triangle for the coefficients (a+b)^2= 1a^2 + 2ab +1b^2= a^2 +2ab+b^2 can you guess whats (a+b)^3? and for (a+b)^4?

  9. mark_o.
    • 3 years ago
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    note it can be like (x+y)^3 ? or (x+y)^4 ?

  10. mark_o.
    • 3 years ago
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    so power of 3 we get the coefficients of 1 3 3 1 therefore the answer for (x+y)^3= x^3 +3x^2y +3 xy^2 + y^3 can you guess for (x+y)^4 ?

  11. mark_o.
    • 3 years ago
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    the reason i want you to learn this is when you go solved for the derivative of y=x^4 you will get to find (x+y)^4

  12. needhelp07
    • 3 years ago
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    tnx a lot . i need to study more about this

  13. mark_o.
    • 3 years ago
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    im here now to teach you the easier way of doing it

  14. mark_o.
    • 3 years ago
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    ok for the power of 4 the coefficents inthe pascal triangle are 1 4 6 4 1 therefore (a+b)^4 = a^4 + 4a^3 b +6 a^2 b^2 +4 a b^3 + b^4 or (x+y)^4= x^4 +4x^3 y + 6 x^2 y^2 + 4x y^3 + y^4 did you see it ?

  15. needhelp07
    • 3 years ago
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    yes

  16. mark_o.
    • 3 years ago
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    now for practice y=x^2 \[y+\Delta y=x +\Delta x\] \[\Delta y=x+\Delta x - y\] \[\Delta y= (x+\Delta x)^{2}-x ^{2}\]

  17. mark_o.
    • 3 years ago
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    can you try solving that first ?

  18. needhelp07
    • 3 years ago
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    wait

  19. needhelp07
    • 3 years ago
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    i try solving it

  20. needhelp07
    • 3 years ago
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    the answer is 2x

  21. mark_o.
    • 3 years ago
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    \[\Delta y=f(x+\Delta x) - y\] \[\Delta y=(x+\Delta x)^{2}- x ^{2}\] \[\Delta y= x ^{2 }+ 2x \Delta x +\Delta x ^{2} -x ^{2}\] add or subtract we get \[\Delta y= 2x \Delta x+\Delta x ^{2}\] now the derivative is \[y'=f'(x)= \lim _{\Delta x ->0}\frac{ \Delta y }{ \Delta x }=\] \[= \lim _{\Delta x->0}(\frac{ 2x \Delta x+\Delta x ^{2} }{ \Delta x })\] try to cancell something in there

  22. needhelp07
    • 3 years ago
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    i know it already the answer is 2x

  23. mark_o.
    • 3 years ago
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    ok good can you try it if its y=x^4 ? do the same process

  24. needhelp07
    • 3 years ago
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    okay

  25. needhelp07
    • 3 years ago
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    the answer is 4x^3+6x^2+4x

  26. mark_o.
    • 3 years ago
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    hmm to make it easier lets make delta x=h so that delta y=(x+h)^4 - x^4 try solving that first

  27. needhelp07
    • 3 years ago
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    I cant follow my professor's teaching. Good to know that u are here to help me .thank you so much. you are a great help! i need to do more assignments about this so i will go out now .

  28. mark_o.
    • 3 years ago
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    the answer should be f'(x)=4x^3

  29. needhelp07
    • 3 years ago
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    haha im wrong

  30. mark_o.
    • 3 years ago
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    do the process i gave you there, and you will arrive at the correct answer ... :D

  31. needhelp07
    • 3 years ago
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    i wish i have a tutor like u hahaha .

  32. mark_o.
    • 3 years ago
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    remember this ?. (x+y)^4= x^4 +4x^3 y + 6 x^2 y^2 + 4x y^3 + y^4 make y an h here so that (x+h)^4= x^4 +4x^3 h + 6 x^2 h^2 + 4x h^3 + h^4

  33. needhelp07
    • 3 years ago
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    anyway thx a lot

  34. needhelp07
    • 3 years ago
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    goodbye!

  35. mark_o.
    • 3 years ago
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    ok just read it through here again and learn from it good luck and have fun now :D

  36. needhelp07
    • 3 years ago
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    ok

  37. mark_o.
    • 3 years ago
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    ok just messege me here and if you have prob i will go and give you an advise or correct if you have something done incorrectly ... :D

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