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