Q. 145.0( 1 Vote )

# Differentiate the

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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