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Differentiate the function with respect to x. (log x)^x + x^log x - Mathematics

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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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अध्याय 5: Continuity and Differentiability - Exercise 5.5 [पृष्ठ १७८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.5 | Q 7 | पृष्ठ १७८

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