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anonymous

  • one year ago

Is this correct? (Spoiler: It's derivatives) http://i.imgur.com/MBKQYup.png

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  1. phi
    • one year ago
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    what is the derivative of \[ 7 x^{-1} \] ?

  2. phi
    • one year ago
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    using the power rule d/dx x^n = n x^(n-1)

  3. anonymous
    • one year ago
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    I wish I could tell you - the content doesn't cover that. It only gives me f(x) and ask me to find it.

  4. phi
    • one year ago
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    The rule is posted above. But here it is again \[ \frac{d}{dx} x^n = n x^{n-1} \]

  5. phi
    • one year ago
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    in your case, n= -1

  6. anonymous
    • one year ago
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    What would variable "d" be then?

  7. anonymous
    • one year ago
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    The \(d\) is notation for an operator. Just like \(+\), it doesn't have a value

  8. phi
    • one year ago
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    d/dx is how you say "derivative with respect to x"

  9. phi
    • one year ago
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    the power rule for finding a derivative is one of the first things you learn in differential calculus. If this is mysterious, you need a refresher, because it is too much material to review here.

  10. anonymous
    • one year ago
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    I'm in Precalc, so we're probably never gonna cover the power rule. This is simply using the difference quotient.

  11. phi
    • one year ago
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    you mean using \[ f'(x) = \lim_{h\rightarrow0}\frac{ f(x+h) - f(x)}{(x+h)-x} \]?

  12. anonymous
    • one year ago
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    Yep, we're doing that to solve for the posted equation.

  13. phi
    • one year ago
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    in that case write \[ f'(x) = \lim_{h\rightarrow0}\frac{\frac{7}{x+h} - \frac{7}{x} }{h} \]

  14. anonymous
    • one year ago
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    Alright, sorry. Internet kinda cut out. I know we gotta find a common denominator between the two fractions.

  15. anonymous
    • one year ago
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    (which would be x for the first fraction, (x+h) for the second if I'm correct in saying that)

  16. phi
    • one year ago
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    or \[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right)\]

  17. phi
    • one year ago
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    multiply the first fraction by x/x and the second by (x+h)/(x+h)

  18. anonymous
    • one year ago
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    \[\frac{ x }{ x^2 +xh} - \frac{ x+h }{ x^2+xh }\] Right?

  19. phi
    • one year ago
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    ok. now combine the top, and put the result over the common denominator

  20. anonymous
    • one year ago
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    \[\frac{ h }{ x^2+xh }\] Alrightie.

  21. phi
    • one year ago
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    the top is x-(x+h) = x-x-h

  22. phi
    • one year ago
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    \[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right) \\ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{-h}{x^2+xh} \right) \]

  23. anonymous
    • one year ago
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    Alright, I see.

  24. phi
    • one year ago
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    notice h/h divides out and you have \[ f'(x) = 7\lim_{h\rightarrow0} \left(\frac{-1}{x^2+xh} \right) \]

  25. phi
    • one year ago
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    now let h "go to zero". the expression approaches -1/(x^2) you get \[ f'(x) = 7 \cdot \frac{-1}{x^2} \] evaluate that at x=1

  26. anonymous
    • one year ago
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    So from here I'd replace x^2 with 1?

  27. phi
    • one year ago
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    you replace x with 1 so you get 1^2 or 1*1 in the denominator.

  28. anonymous
    • one year ago
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    Which gives the end result of -7?

  29. phi
    • one year ago
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    yes

  30. anonymous
    • one year ago
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    Alright, thank you so much for your help.

  31. phi
    • one year ago
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    yw

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