Question

Differentiate x^x w.r.t. x^six.

Answer 1

Let u = x^x and v = x^sinx
Then we want to find `"du"/"dx"`.
Take, u = x^x
∴ 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)`
= x^x(1 + log x)
Also, v = x^sinx
∴ log v = logx^sinx = (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)`.