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amy0799

  • one year ago

if f(x) = (x^2-c^2)/(x^2+c^2) where c is a constant, find f'(x)

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  1. IrishBoy123
    • one year ago
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    \[f(x) = \frac {x^2-c^2}{x^2+c^2}\] simplify? \[f(x) = \frac {x^2+c^2-2c^2}{x^2+c^2} = 1 - \frac{2c^2}{x^2 + c^2}\] can you finish this?

  2. amy0799
    • one year ago
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    i thought it would be \[\frac{ 2x-2c }{ 2x+2c }\]

  3. IrishBoy123
    • one year ago
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    why did you think that?

  4. amy0799
    • one year ago
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    hold on, i know what i did wrong

  5. IrishBoy123
    • one year ago
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    and remember, c is a constant so \(\frac{d}{dx} \left[ c^2 \right] = 0\)

  6. amy0799
    • one year ago
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    \[\frac{ (x ^{2} +c ^{2})2x-(x ^{2{}}-c ^{2})2x}{ (x ^{2} +c ^{2})^{2}}\] is this right?

  7. IrishBoy123
    • one year ago
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    you are applying the quotient rule. before i look at your work, is that what you are supposed to be doing with this because it is a silly way to do it. it just adds complications. let me know either way and we can proceed

  8. IrishBoy123
    • one year ago
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    your application of the quotient rule is correct

  9. IrishBoy123
    • one year ago
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    if you wish to do it this way, next step is to simplify the numerator

  10. amy0799
    • one year ago
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    simpliflying it would get me \[\frac{ 4xc ^{2} }{ (x ^{2}+c ^{2})^{2} }\]

  11. IrishBoy123
    • one year ago
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    well done!

  12. amy0799
    • one year ago
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    so that's the answer?

  13. IrishBoy123
    • one year ago
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    yes

  14. amy0799
    • one year ago
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    thank you!

  15. IrishBoy123
    • one year ago
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    mp

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