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# differentiation p3 q watch

1. prove:

d/dx (1+ cot x)/(1 - cot x) = 2/(sin2x + 1)
2. whoops done it... differentiated cot x as -cosec x rather than -cosec^2 x
3. (Original post by powerball)
prove:

d/dx (1+ cot x)/(1 - cot x) = 2/(sin2x + 1)
d/dx=[-(1-cotx)cosec²x-(1+cotx)cosec²x]/(1-cotx)²
=[-2cosec²x]/(1-cotx)²

cotx=cosx/sinx
So 1-cotx=1-cosx/sinx
=(sinx-cosx)/sinx

Thus dy/dx=[-2/sin²x]/[(sinx-cosx)²/sin²x]
=-2/(sinx-cosx)²
=-2/(1-sin2x)
=2/(sin2x-1)

I think the question's wrong, as the answer's definately 2/(sin2x-1)
4. (1 + cot(x)) / (1 - cot(x))
= (sin(x) + cos(x)) / (sin(x) - cos(x)).

d/dx --->

[(sin(x) - cos(x))(cos(x) - sin(x)) - (sin(x) + cos(x))^2] / (sin(x) - cos(x))^2
= -2(sin(x)^2 + cos(x)^2) / (sin(x) - cos(x))^2
= -2 / (sin(x) - cos(x))^2
= -2 / (1 - 2sin(x)cos(x))
= 2 / (sin(2x) - 1).

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Updated: June 9, 2004
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