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Littlebird

  • one year ago

Difference Quotient

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  1. Littlebird
    • one year ago
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    f(x) = 1/x^4

  2. Littlebird
    • one year ago
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    I've managed to get to (-4x^3-6hx^2-4h^2x-h^3)/(x^4*(x+h)^4)

  3. triciaal
    • one year ago
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    what do you mean by difference quotient? difference what you get when you subtract quotient what you get when you divide

  4. triciaal
    • one year ago
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    |dw:1443157711272:dw|

  5. Littlebird
    • one year ago
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    (f(x+h)-f(x))/h

  6. gibbs
    • one year ago
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    F

  7. misty1212
    • one year ago
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    are you trying to take the derivative by hand ?

  8. Littlebird
    • one year ago
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    Yes

  9. Littlebird
    • one year ago
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    I know the shortcut

  10. misty1212
    • one year ago
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    then you need to compute \[\lim_{h\to 0}\frac{1}{(x+h)^4}-\frac{1}{x^4}\] right?

  11. gibbs
    • one year ago
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    1 Factor out the common term f f(x+h−x)/h 2 Gather like terms f((x−x)+h)/h 3 Simplify (x−x)+h to h fh/h 4. Cancel h f

  12. ytrewqmiswi
    • one year ago
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    u mean the shortcut to find the nth derivative :)

  13. misty1212
    • one year ago
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    someone needs to stop @gibbs who clearly does not know what he is talking about

  14. Littlebird
    • one year ago
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    yes misty, you have what I wrote, but I also divided everything with h

  15. Jhannybean
    • one year ago
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    Difference Quotient: \(\lim_{h\rightarrow 0} \dfrac{f(x+h) - f(x)}{h}\)

  16. ytrewqmiswi
    • one year ago
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    lml he used this XD - http://prntscr.com/8kb4j9

  17. Jhannybean
    • one year ago
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    -_____-

  18. Jhannybean
    • one year ago
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    So find all the hard stuff first: \(f(x+h) = \dfrac{1}{(x+h)^4}\) put it back into the formula : \(\dfrac{\dfrac{1}{(x+h)^4} - \dfrac{1}{x^4}}{h}\) Then start simplifying by finding a common denominator in the numerator portion.

  19. misty1212
    • one year ago
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    ok leave out the \(h\) in the denominator for a moment and focus just on \[\frac{1}{(x+h)^4}-\frac{1}{x^4}\] we will divide by ]\(h\) at the end

  20. misty1212
    • one year ago
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    it is now 100% algebra \[\frac{x^4-(x+4)^4}{x^4(x+h)^4}\]

  21. misty1212
    • one year ago
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    you have no choice here but to expand in the numerator sorry

  22. Littlebird
    • one year ago
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    I just noticed that misty put a limit of h->0 next to the thing before If I use that I actually get the answer

  23. Jhannybean
    • one year ago
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    Ugh binomial expansion.

  24. misty1212
    • one year ago
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    no choice

  25. Jhannybean
    • one year ago
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    NO choice. you are doomed.

  26. Littlebird
    • one year ago
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    (-4x^3-6hx^2-4h^2x-h^3)/(x^4*(x+h)^4)

  27. misty1212
    • one year ago
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    ooh k

  28. Littlebird
    • one year ago
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    am I normally supposed to have a limit when solving these problems?

  29. misty1212
    • one year ago
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    looks like you cancel the \(h\) to right? so now all that is left is to replace \(h\) by \(0\) and you are done

  30. misty1212
    • one year ago
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    yes, if you are going to do it by hand

  31. Jhannybean
    • one year ago
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    \[\dfrac{x^4-x^4-16x^3 - 96x^2-256x -256}{x^4(x+4)^4}\]

  32. Jhannybean
    • one year ago
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    It's just too much to constantly write the limit over and over again in latex, so i just leave it for the end... it's wrong but <_<

  33. Jhannybean
    • one year ago
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    And then you simplify the denominator there at the bottom

  34. Littlebird
    • one year ago
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    I already had it expanded. the only thing i was missing was the limit

  35. Littlebird
    • one year ago
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    you dont have to do all of this ;)

  36. Littlebird
    • one year ago
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    but thankyou

  37. Jhannybean
    • one year ago
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    Why have you multiplied h to the numerator portion? \(\color{#0cbb34}{\text{Originally Posted by}}\) @Littlebird \(\color{red}{(-4x^3-6hx^2-4h^2x-h^3)}/(x^4*(x+h)^4) \(\color{#0cbb34}{\text{End of Quote}}\) That portion.

  38. Littlebird
    • one year ago
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    I multiplied both the top and bottom of the fraction by x^4(x+h)^4 so that I could make the fractions on top go away

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