Question

The derivative of y=logx^x is.Pls explain with full solution…

Answer 1

Answer:

y=(logx)x

Taking log on both sides, we get

logy=log(logx)x

logy=xlog(logx)

y1.dxdy=x×logx1×x1+log(logx)

y1dxdy=logx1+log(logx)

dxdy=(logx)x[logx1+log(logx)]

hope it helps

Answer 2

Step-by-step explanation:

y=(logx)^x

Taking log on both sides, we get

logy=log(logx)^x

logy=xlog(logx)

1/y•dy/dx= x × 1/logx × 1/x+log(logx)

1/y dy/dx= 1/logx + log(logx)

dy/dx= (logx)^x [1/logx + log(logx)]

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