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