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))`

Answer 1

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