Find dydx, if y = xxx - Mathematics and Statistics

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Sum

Find `("d"y)/("d"x)`, if y = `x^(x^x)`

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Solution

y = `x^(x^x)`

Taking logarithm of both sides, we get

log y = `log x^(x^x)`

∴ log y = xx log x

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

`"d"/("d"x)(log y) = x^x*"d"/("d"x)(log x) + logx*"d"/("d"x)(x^x)`

∴ `1/y*("d"y)/("d"x) = x^x*1/x + logx*"d"/("d"x)(x^x)`  ......(i)

Let u = xx

Taking logarithm of both sides, we get

log u = log xx

∴ log u = x log x

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

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

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

∴ `1/"u"*"du"/("d"x)` = 1 + log x

∴ `"du"/("d"x)` = u(1 + log x)

∴ `"d"/("d"x)(x^x)` = xx(1 + log x)    ......(ii)

Substituting (ii) in (i), we get

`1/y*("d"y)/("d"x) = x^x*1/x + logx*x^x(1 + log x)`

∴ `("d"y)/("d"x) = yx^x[1/x + logx(1 + logx)]`

∴ `("d"y)/("d"x) = x^(x^x)*x^x[1/x + logx(1 + logx)]`

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

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