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प्रश्न
Differentiate the following w.r.t. x :
etanx + (logx)tanx
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उत्तर
Let y = etanx + (logx)tanx
Put u = (logx)tanx
∴ log u = log (log x)tanx = (tan x).(log log x)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[(tan x).(log logx)]`
= `(tanx)."d"/"dx"(log log x) + (log log x)."d"/"dx"(tan x)`
= `tanx xx 1/logx."d"/"dx"(logx) + (log log x)(sec^2x)`
= `tanx xx 1/logx xx 1/x + (log log x)(sec^2x)`
∴ `"du"/"dx" = u[tanx/(xlogx) + (loglogx)(sec^2x)]`
= `(logx)^(tanx)[tanx(xlogx) + (log log x)(sec^2x)]`
Now, y = etanx + u
∴ `"dy"/"dx" = "d"/"dx"(e^tanx) + "du"/"dx"`
= `e^(tanx)."d"/"dx"(tanx) + "du"/"dx"`
= `e^(tanx).sec^2x + (logx)^(tanx)[tanx/(xlogx) + (log log x)(sec^2x)]`
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