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lukmanshafa

  • 2 years ago

find d/dx of f^(-1) (x)

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  1. ezhyl_marie
    • 2 years ago
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    what is the value of x?

  2. lukmanshafa
    • 2 years ago
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    x/x + 1

  3. funinabox
    • 2 years ago
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    is f(x) = x/x + 1? or is f^-1(x)= x/x +1?

  4. lukmanshafa
    • 2 years ago
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    d 2

  5. lukmanshafa
    • 2 years ago
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    f^(-1) x = x/x+1

  6. funinabox
    • 2 years ago
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    well in any case, use the quotient rule where if you a function h defined as f/g then d/dx h = (f' * g - f * g') / g^2

  7. ezhyl_marie
    • 2 years ago
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    ..............x f(x) = ------------- ..........x + 1 ........x y = -------- ......x + 1 ........y x = -------- ......y + 1 x(y + 1) = y xy + x = y xy - y = - x y(x - 1) = - x ...........x y = - --------- .........x - 1 .................x f⁻¹(x) = - ----------- answer// ................x - 1 bye..

  8. lukmanshafa
    • 2 years ago
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    nope

  9. lukmanshafa
    • 2 years ago
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    i got dat but it wasn't d ans

  10. funinabox
    • 2 years ago
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    you need to use the quotient rule next: x = f, x+1 = g d/dx x/x+1 = [1(x+1) - x(1)]/(x+1)^2 1/(x+1)^2

  11. lukmanshafa
    • 2 years ago
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    yeah i got it now but it wasnt ur ans funinabox it was 1/(1-x)^2

  12. lukmanshafa
    • 2 years ago
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    thank u all

  13. funinabox
    • 2 years ago
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    then you typed the problem wrong

  14. lukmanshafa
    • 2 years ago
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    no it was correct

  15. funinabox
    • 2 years ago
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    then the answer is wrong 1/(1+x)^2 is the correct answer.

  16. lukmanshafa
    • 2 years ago
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    but i got the marks

  17. funinabox
    • 2 years ago
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    i think i know what's up f(x) = x/x + 1 it's not f^-1(x) = x/x + 1 as was stated that way, when you do find the inverse, you get x/1-x, which would give you the derivative you posted.

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