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Question
Differentiate the function with respect to x.
(log x)x + xlog x
Solution
Let, y = (log x)x + xlog x
Again, let y = u + v
Differentiating both sides with respect to x,
`(dy)/dx = (du)/dx + (dv)/dx` ....(1)
अब, u = (log x)x
Taking logarithm of both sides,
log v = log (log x)x = x log (log x) ...[∵ log mn = n log m]
Differentiating both sides with respect to x,
`1/u (du)/dx = x d/dx log (log x) + log (log x) d/dx (x)`
`= x * 1/(log x) d/dx (log x) + log (log x) xx 1`
`= x * 1/(log x) 1/x + log (log x) = 1/(log x) + log (log x)`
`therefore (du)/dx = u [log (log x) + 1/(log x)] = (log x)^x [log (log x) + 1/log x]`
तथा v = `x^(log x)`
Taking logarithm of both sides,
log v = log xlog x = log x log x = (log x)2
Differentiating both sides with respect to x,
`1/v (dv)/dx = d/dx (log x)^2 = 2 log x d/dx log x = (2 log x)/x`
`therefore dv/dx = v (2/x log x) = 2/x (x^(log x) log x)`
From equation (1),
`(dy)/dx = (du)/dx + (dv)/dx`
`∴ dy/dx = (logx^x) [1/logx + (logx)] + x^(log x) [(2 log x)/x]`
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