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प्रश्न
Differentiate the function with respect to x.
(log x)x + xlog x
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उत्तर
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)
Now, u = (log x)x
Taking logarithm of both sides,
log u = log (log x)x ...[∵ log mn = n log m]
log u = x log (log x)
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)`
= `u [log (log x) + 1/(log x)]`
∴ `(du)/dx = (log x)^x [log (log x) + 1/log x]`
Also v = xlog 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 xx 1/x`
= `v ((2 log x)/x)`
∴ `(dv)/dx = x^(log x)((2 log x)/x)`
From equation (1),
`(dy)/dx = (du)/dx + (dv)/dx`
`∴ dy/dx = (log x)^x [log (log x) + 1/log x] + x^(log x) ((2 log x)/x)`
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