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Question:

Differentiate the following function with respect to x: (logx)x + xlogx

Solution:

Let y = (logx)x + xlogx
Then dy/dx = d((logx)x + xlogx)/dx = d((logx)x)/dx + d(xlogx)/dx
= (logx)x d(xlog(logx))/dx + x(logx) d(logx)/dx
From (d(uv)/dx = u(dv/dx) + v(du/dx))
= (logx)x (1/logx + log(logx)) + xlogx (1/x)
Therefore, d((logx)x + xlogx)/dx = (logx)x (1/logx + log(logx)) + xlogx(1/x)