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Question
Find the derivatives of the following:
Find the derivative with `tan^-1 ((sinx)/(1 + cos x))` with respect to `tan^-1 ((cosx)/(1 + sinx))`
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Solution
Let u = `tan^-1 ((sinx)/(1 + cos x))`
= `tan^-1 ((2 sin x/2 cos x/2)/(2 cos ^2 x/2))`
= `tan^-1 ((sin x/2)/(cos x/2))`
= `tan^-1 (tan x/2)`
u = `x/2`
`("d"u)/("d"x) = 1/2` .......(1)
v = `tan^-1 [cos x/(1 + sin x)]`
= `tan^-1 [(cos^2 x/2 - sin^2 x/2)/(cos^2 x/2 + sin^2 x/2 + 2 sin x/2 cos x/2)]`
= `tan^-1 [((cos x/2 + sin x/2)(cos x/2 - sin x/2))/(cos x/2 + sin x/2)^2]`
= `tan^-1 [(cos x/2 - sin x/2)/(cos x/2 + sin x/2)]`
= `tan^-1 [(cos x/2 (1 - (sin x/2)/(cos x/2)))/(cos x/2 (1 + (sin x/2)/(cos x/2)))]`
= `tan^-1 [(1 - tan x/2)/(1 + tan x/2)]`
= `tan^-1 [(tan pi/4 - tan pi/2)/(1 + tan pi/4 * tan x/2)]`
= `tan^-1 [tan (pi/4 - x/2)]`
v = `pi/4 - x/2`
`("d"v)/("d"x) = - 1/2` .........(2)
From equations (1) and (2)
`(("d"u)/("d"x))/(("d"v)/("d"x)) = (1/2)/(- 1/2)`
`("d"u)/("d"v)` = – 1
`("d"tan^-1 [(sin x)/(1 + cos x)])/("d"tan^-1 [cos x/(1 + sin x)])` = – 1
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