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Q.

If y=logxx then dydx is

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a

(logx)xlog(logx)+1logx

b

(logx)xlog(logx)−1logx

c

−(logx)xlog(logx)+1logx

d

−(logx)xlog(logx)−1logx

answer is A.

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Detailed Solution

Given y=logxxTake logarithm on both sideslogy=loglogxxlog y =x ·log(logx)Differentiate both sides 1ydydx=loglogx+x1logxddxlogx=loglogx+xlogx1x Therefore, dydx=yloglogx+1logx=logxxloglogx+1logx

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