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Question
Find the derivatives of the following:
x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)`
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Solution
x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)`
Put t = tan θ
x = `(1 - tan^2theta)/(1 + tan^2theta)` y = `(2tantheta)/(1 + tan^2theta)`
x = cos 2θ, y = sin 2θ
`("d"x)/("d"theta) = - sin 2theta xx 2`, `("d"y)/("d"theta) = cos 2theta xx 2`
`("d"x)/("d"theta) = - 2 sin 2theta`, `("d"y)/("d"theta) = 2 cos 2theta`
`(("d"y)/("d"theta))/(("d"x)/("d"theta)) = (2 cos 2theta)/(-2 sin 2theta)`
`("d"y)/("d"x) = - cot 2theta`
tan 2θ = `(2tantheta)/(1 - tan^2theta)`
cot 2θ = `(1 - tan^2theta)/(2tantheta)`
= `- (1 - tan^2theta)/(2tantheta)`
= `- (1 - "t"^2)/(2"t")`
`("d"y)/("d"x) = ("t"^2 - 1)/(2"t")`
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