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प्रश्न
If y = (log x)x + xlog x, find `"dy"/"dx".`
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उत्तर
Let y =(log x)x + xlog x
Also, let u =(log x)x and v = xlog x
∴ y = u + v
`⇒"dy"/"dx" = "du"/"dx"+"dv"/"dx"` ........(1)
u = (logx)x
⇒ log u = log[(log x)x]
⇒ log u = x log(log x)
Differentiating both sides with respect to x, we obtain
`1/"u" "du"/"dx" ="d"/"dx"("x") xx log(log"x")+"x"."d"/"dx"[log(log"x")]`
`⇒"du"/"dx" = "u"[1xxlog(log"x")+"x". 1/log"x"."d"/"dx"(log"x")]`
`⇒"du"/"dx"=(log"x")^"x"[log(log"x")+"x"/(log"x"). 1/"x"]`
`⇒"du"/"dx"=(log"x")^"x"[log(log"x")+1/(log"x")]`
`⇒"du"/"dx"=(log"x")^"x"[(log(log"x").log"x"+1)/log"x"]`
`⇒"du"/"dx"=(log"x")^("x"-1)[1+log"x".log(log"x")]` .....(2)
v = xlogx
⇒ log v = log(xlogx)
⇒ log v = log x log x = (log x)2
Differentiating both sides with respect to x, we obtain
`1/"v"."dv"/"dx"="d"/"dx"[(log"x")^2]`
`⇒ 1/"v"."dv"/"dx"=2(log"x")."d"/"dx"(log"x")`
`⇒"dv"/"dx" = 2"v"(log"x"). 1/"x"`
`⇒"dv"/"dx" = 2"x"^(log"x") log"x"/"x"`
`⇒"dv"/"dx"=2"x"^(log"x"-1) . log"x"` ......(3)
Therefore, from (1), (2), and (3), we obtain
`"dy"/"dx" = (log"x")^("x"-1)[1+log"x".log(log"x")]+2"x"^(log"x"-1) .log"x"`
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