Question

Differentiate `tan^-1((cosx)/(1 + sinx)) w.r.t. sec^-1 x.`

Answer 1

Let u = `tan^-1((cosx)/(1 + sinx)) and v = sec^-1x.`
Then we want to find `"du"/"dv"`.
Differentiate u and v w.r.t. x, we get
`"du"/"dx" = "d"/"dx"[tan^-1((cos)/(1 + sinx))]`

`(cosx)/(1 + sinx) = (sin(pi/2 - x))/(1 + cos(pi/2 - x)`

= `(2sin(pi/4 - x/2).cos(pi/4 - x/2))/(2cos^2(pi/4 - x/2)`

= `tan(pi/4 - x/2)`

∴ `"du"/"dx" = "d"/"dx"[tan^-1 tan(pi/4 - pi/2)]`

= `"d"/"dx"(pi/4 - x/2)`

= `"d"/"dx"(pi/4) - (1)/(2)"d"/"dx"(x)`

We know `d/dx(x) = 1` and the derivative of a constant is 0.

= `0 - (1)/(2) xx 1`

= `-(1)/(2)`
and
`"dv"/"dx" = "d"/"dx"(sec^-1x)`

= `(1)/(xsqrt(x^2 - 1)`

∴ `"du"/"dv" = (("du"/"dx"))/(("dv"/"dx")`

= `((-1/2))/((1/(xsqrt(x^2 - 1)))`

= `-(xsqrt(x^2 - 1))/(2)`.