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

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what is the value of x?
x/x + 1
is f(x) = x/x + 1? or is f^-1(x)= x/x +1?

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Other answers:

d 2
f^(-1) x = x/x+1
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
..............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..
i got dat but it wasn't d ans
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
yeah i got it now but it wasnt ur ans funinabox it was 1/(1-x)^2
thank u all
then you typed the problem wrong
no it was correct
then the answer is wrong 1/(1+x)^2 is the correct answer.
but i got the marks
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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