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Find the derivatives of the following:y = xlogx+(logx)x - Mathematics

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प्रश्न

Find the derivatives of the following:
y = `x^(logx) + (logx)^x`

बेरीज
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उत्तर

y = `x^(logx) + (logx)^x`

Let u = xlog x, v = (log x)x

log u = log xlog x

log u = (log x)(log x)

log u = (log x)2 

`1/u * ("d"u)/("d"x) = 2 log x xx 1/x`

`("d"u)/("d"x) = 2u (logx)/x`  ........(1)

v = (log x)x

log v = log (log x)x

log v = x log (log x)

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

`("dv")/("d"x) = "v"[1/logx + log(logx)]` ........(2)

y = u + v

`("d"y)/("d"x) = ("d"u)/("d"x) + ("dv")/("d"x)`

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

By equation (1) and (2)

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

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Differentiation Rules
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Differential Calculus - Differentiability and Methods of Differentiation - Exercise 10.4 [पृष्ठ १७६]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.4 | Q 2 | पृष्ठ १७६

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