Find dydx, if y = (log x)x + (x)logx - Mathematics and Statistics

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Sum

Find `("d"y)/("d"x)`, if y = (log x)x + (x)logx

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Solution

y = (log x)x + (x)log

Let u = (log x)x and v = xlogx 

∴ y = u + v

Differentiating both sides w. r. t. x, we get

`("d"y)/("d"x) = "du"/("d"x) + "dv"/("d"x)`   .....(i)

Now, u = (log x)x

Taking logarithm of both sides, we get

log u = log (log x)x = x log (log x)

Differentiating both sides w. r. t. x, we get

`"d"/("d"x)(log "u") = x*"d"/("d"x)[log(logx)] + log(logx)*"d"/("d"x)(x)`

∴ `1/"u"."du"/("d"x) = x*1/(logx)*"d"/("d"x)(log x) + log(logx)*1`

∴ `1/"u"*"du"/("d"x) = x*1/(logx)*1/x + log(log x)`

∴ `"du"/("d"x) = "u"[1/logx + log(logx)]`

∴ `"du"/("d"x) = (log x)^x [1/logx + log(log x)]`  .....(ii)

Also, v = xlogx

Taking logarithm of both sides, we get

log v = log (xlogx) = log x (log x)

∴ log v = (log x)2 

Differentiating both sides w.r.t. x, we get

`1/"v"*"dv"/("d"x) = 2logx*"d"/("d"x)(log x)`

∴ `1/"v"*"dv"/("d"x) = 2logx*1/x`

Substituting (ii) and (iii) in (i), we get∴ `"d"/("d"x) = "v"[(2logx)/x]`

∴ `"dv"/("d"x) = x^(logx)[(2logx)/x]`   ......(iii)

Substituting (ii) and (iii) in (i), we get

`("d"y)/("d"x) = (log x)^x[1/logx + log(logx)] + x^(logx)[(2logx)/x]`

Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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Chapter 1.3: Differentiation - Q.5

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