Question

Differentiate log `[(sqrt(1 + x^2) + x)/(sqrt(1 + x^2 - x)]]` w.r.t. cos (log x).

Answer 1

Let y = log `[(sqrt(1 + x^2) + x)/(sqrt(1 + x^2 - x)]]` and v = cos (log x)

Then we want to find `"du"/"dv"`.

u = `log((sqrt(1 + x^2) + x)/(sqrt(1 + x^2 - x)) xx (sqrt(1 + x^2) + x)/(sqrt(1 + x^2 + x)))`

= `log[((sqrt(1 + x^2) + x)^2)/(1 + x^2 - x^2)]`

= `2 log (sqrt(1 + x^2) + x)`

∴ `"du"/"dx" = 2"d"/"dx"[log(sqrt(1 + x^2) + x)]`

= `(2)/(sqrt(1 + x^2) + x)."d"/"dx"(sqrt(1 + x^2) + x)`

= `(2)/(sqrt(1 + x^2) + x).[1/(2sqrt(1 + x^2))."d"/"dx"(1 + x^2) + 1]`

= `(2)/(sqrt(1 + x^2) + x).[(2x)/(2sqrt(1 + x^2)) + 1]`

= `(2)/(sqrt(1 + x^2) + x)(x/sqrt(1 + x^2) + 1)`

= `(2(x + sqrt(1 + x^2)))/((sqrt(1 + x^2) + x)sqrt(1 + x^2)`

= `(2)/sqrt(1 + x^2)`

`"dv"/"dx" = "d"/"dx"[cos(logx)]`

= `-sin(logx)"d"/"dx"(logx)`

= `[-sin(logx)] xx (1)/x`

= `(-sin(logx))/x`

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

= `(((2)/(sqrt(1 + x^2))))/[[((-sin(logx)))/"x"]`

= `(-2x)/(sqrt(1 + x^2).sin(logx))`.