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anonymous

  • one year ago

Find the derivative of f(x) = 6/x at x = -2.

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  1. anonymous
    • one year ago
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    solomon help haha

  2. SolomonZelman
    • one year ago
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    Ok, lets re-write the f(x) \(\large\color{black}{ \displaystyle f(x)=6(x)^{-1} }\)

  3. SolomonZelman
    • one year ago
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    Apply the power rule. Can you do that?

  4. SolomonZelman
    • one year ago
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    (YOu are to find the derivative, and then plug in x=-2 into the derivative)

  5. anonymous
    • one year ago
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    -3

  6. SolomonZelman
    • one year ago
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    is that your fnal answer?

  7. SolomonZelman
    • one year ago
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    if so, then you are not correct....

  8. SolomonZelman
    • one year ago
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    Did you find the \(f'(x)\) /?

  9. SolomonZelman
    • one year ago
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    Oh, what I mean by the power rule is: \(\large\color{black}{ \displaystyle \frac{d }{dx}x^n=nx^{n-1} }\) where d/dx is jst a notation for taking the derivative. ------------------------------------------------ But I guess you are doing by the first principles...

  10. anonymous
    • one year ago
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    'm not really sure what you mean by power rule I have a formula for difference quotient f(h-1)-f(1)/h and I ended up with ([6/h-1]-6)/h

  11. anonymous
    • one year ago
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    never heard of it this is precalc

  12. SolomonZelman
    • one year ago
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    yes, you are applying the following: \(\large\color{black}{ \displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)

  13. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)

  14. SolomonZelman
    • one year ago
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    YOu can use the power rule I posted, to at least check the work, but for now I guess we need this: \(\large\color{black}{ \displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)

  15. SolomonZelman
    • one year ago
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    Or, if you want to find \(f'(a)\) direclty: \(\large\color{black}{ \displaystyle f'(a)= \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} }\)

  16. anonymous
    • one year ago
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    (6/x+h)-(6/x)/h

  17. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6}{x+h}-\dfrac{6}{x} }{h} }\)

  18. SolomonZelman
    • one year ago
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    that is right.

  19. SolomonZelman
    • one year ago
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    now, find the common denominator betwen 6/(x+h) and 6/x and subtract.

  20. anonymous
    • one year ago
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    so would I multiply one side by x and the other by x+h

  21. SolomonZelman
    • one year ago
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    yes, fraction#1 by x on top and bottom, and fraction#2 (x+h) on top and bottom

  22. anonymous
    • one year ago
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    I got -6

  23. anonymous
    • one year ago
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    pretty sure that's it thanks

  24. SolomonZelman
    • one year ago
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    no it is not it

  25. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6}{x+h}-\dfrac{6}{x} }{h} }\) \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6x}{x(x+h)}-\dfrac{6(x+h)}{x(x+h)} }{h} }\)

  26. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6x-6(x+h)}{x(x+h)} }{h} }\)

  27. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{-6h}{x(x+h)} }{h} }\)

  28. anonymous
    • one year ago
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    yep that's what I got just forgot about the denomenator under the -6h

  29. SolomonZelman
    • one year ago
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    then divide top and bottom by h.

  30. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{-6h}{x(x+h)}\color{red}{\div h} }{h \color{red}{\div h}} }\)

  31. SolomonZelman
    • one year ago
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    then you get: \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{-6}{x(x+h)} }\)

  32. anonymous
    • one year ago
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    -6/x(x+h)

  33. SolomonZelman
    • one year ago
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    yes, but you are leaving out that important limit.

  34. anonymous
    • one year ago
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    so I just put -2 in for x

  35. SolomonZelman
    • one year ago
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    that limit that h=0, is an important component. So that when you simplify the expression, you then plug in h=0 (if you don;t get any undefined results for that)

  36. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{-6}{x(x+h)} =\frac{-6}{x(x+0)}=\frac{-6}{x^2} }\)

  37. SolomonZelman
    • one year ago
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    see what is that limit for? (it is a notation for the fact that h is 0)

  38. anonymous
    • one year ago
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    - 3/2

  39. SolomonZelman
    • one year ago
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    yes, that is right:

  40. anonymous
    • one year ago
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    hey, thanks for the patience I'm kind of slow

  41. SolomonZelman
    • one year ago
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    Yes, don't forget that limit h->0 notation. it is important.

  42. SolomonZelman
    • one year ago
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    So just an addition that in general: \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{f(x+h)-f(x)}{h} }\) (Derivative a function f(x).) \(\large\color{black}{ \displaystyle f'(x)= \lim_{x \rightarrow a}{~} \frac{f(x)-f(a)}{x-a} }\) (Derivative a function f(x) evaluated at x=a.)

  43. SolomonZelman
    • one year ago
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    good luck

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