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प्रश्न
Differentiate the following w.r.t. x : (logx)x – (cos x)cotx
उत्तर
Let y = (log x)x – (cos x)cotx
Put u = (log x)x and v = (cos x)cotx
Then y = u – v
∴ `"dy"/"dx" = "du"/"dx" - "dv"/"dx"` ...(1)
Take u = (log x)x
∴ log u = log(log x)x = x.log(log x)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[x.log(logx)]`
= `x"d"/"dx"[log(logx)] + log(logx)."d"/"dx"(x)`
= `x xx 1/logx."d"/"dx"(logx) + log(logx) xx 1`
= `x xx 1/logx xx 1/x + log(logx)`
∴ `"du"/"dx" = u[1/logx + log(logx)]`
= `(logx)^x[1/logx + log(logx)]` ...(2)
Also v = (cos x)cotx
∴ log v = log(cos x)cotx = (cot x).(log cos x)
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/dx"[(cotx).log(cosx)]`
= `(cotx)."d"/"dx"(log cosx) + (log cosx)."d"/"dx"(cotx)`
= `cotx xx 1/cosx."d"/"dx"(cosx) + (logcosx)(-"cosec"^2x)`
= `cotx xx 1/cosx xx (-sinx) - ("cosec"^2x)(logcosx)`
∴ `"dv"/"dx" = v[1/tanx xx (-tanx) - ("cosec"^2x)(logcosx)]`
= –(cos x)cotx[1 + (cosec2x)(log cos x)] ....(3)
From (1), (2) and (3), we get
∴ `"dy"/"dx" = (logx)^x[1/logx + log(logx)] + (cosx)^cotx[1 + ("cosec"^2x)(logcosx)]`.
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