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