If y=(logx)x then dydx=

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(logx)xlog(logx)+1logx

(logx)xlog(logx)−1logx

−(logx)xlog(logx)+1logx

−(logx)xlog(logx)−1logx

answer is A.

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Detailed Solution

y=(logx)xlogy=log(logx)xlogy=xlog(logx)1ydydx=log(logx)+x1logxddxlogx=log(logx)+xlogx1xdydx=ylog(logx)+1logx