Q. 145.0( 1 Vote )

Differentiate the

Answer :

Let y = (log x)x + xlog x …(i)

Now, let p = (log)x and q = xlog x


y = p + q …(ii)


Differentiating the above equation with respect to x, we get


…(A)


Now, consider p = (log x)x


Firstly, taking log on both the sides, we get


log p = log[(log x)x]


log p = x log(log x)


Differentiating with respect to x, we get







…(iii)


Now, consider q = xlog x


Taking log on both the sides, we get


log q = log (xlog x)


log q = (log x)(log x)


log q = (log x)2


Differentiating with respect to x, we get








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