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Question
Differentiate xx w.r.t. xsix.
Solution
Let u = xx and v = xsinx
Then we want to find `"du"/"dx"`.
Take, u = xx
∴ log u = x log x
Differentiating both sides w.r.t. x, we get
`(1)/u."du"/"dx" = "d"/"dx"(xlogx)`
= `x"d"/"dx"(logx) + (logx)."d"/"dx"(x)`
= `x xx (1)/x + (logx) xx 1`
∴ `"du"/"dx" = u(1 + logx)`
= xx(1 + log x)
Also, v = xsinx
∴ log v = logxsinx = (sin x)(log x)
Differentiating both sides w.r.t. x, we get
`(1)/v."dv"/"dx" = "d"/"dx"[(sin x)(logx)]`
= `(sinx)."d"/"dx"(logx) + (logx)."d"/"dx"(sinx)`
= `(sinx) xx (1)/x + (logx)(cosx)`
∴ `"dv"/"dx" = v[sinx/x + (logx)(cosx)]`
= `x^(sinx)[sinx/x + (logx)(cosx)]`
∴ `"du"/"dv" = (("du"/"dx"))/(("dv"/"dx")`
= `(x^x(1 + logx))/(x^(sinx)[sinx/x + (logx)(cosx)]`
= `(x^x(1 + log x) xx x)/(x^(sinx)[sinx + x cosx.logx]`
= `((1 + logx).x^(x+ 1 - sinx))/(sinx + xcosx.logx)`.
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