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If y=(log⁡x)x then dydx=

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a

(log⁡x)xlog⁡(log⁡x)+1log⁡x

b

(log⁡x)xlog⁡(log⁡x)−1log⁡x

c

−(log⁡x)xlog⁡(log⁡x)+1log⁡x

d

−(log⁡x)xlog⁡(log⁡x)−1log⁡x

answer is A.

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Detailed Solution

y=(log⁡x)xTake logarithm on both sideslog⁡y=log⁡(log⁡x)xDifferentiate both sides 1ydydx=loglogx+x1logxddxlogx1ydydx=loglogx+x1logxddxlogx=loglogx+xlogx1x\ Therefore, dydx=ylog⁡(log⁡x)+1log⁡x=(log⁡x)xlog⁡(log⁡x)+1log⁡x

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If y=(log⁡x)x then dydx=