How do you differentiate the following: #v=x^x#?
My steps:
1. #v=x^x#
2. #x^x = e^ln(x^x) = e^(x lnx)#
3. #(e^(x lnx))' = (e^(x lnx))lnx#
4. #(dv)/dx = (x^x)lnx#
Correct Answer: #(dv)/dx = x^x(1+lnx)#
My steps:
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Correct Answer:
1 Answer
Jul 28, 2017
# (dv)/dx = x^x(1 + lnx) #
Explanation:
My approach would be to use implicit logarithm differentiation:
We have:
# v=x^x #
Taking Natural logarithms we have:
# lnv = ln {x^x} #
# " " = xlnx #
Implicitly differentiating and applying the product rule we get:
# d/dx lnv = (x)(d/dx ln x) + (d/dx x)(lnx) #
# :. 1/v (dv)/dx = x 1/x + 1 lnx #
# " " = 1 + lnx #
# :. (dv)/dx = v(1 + lnx) #
# " " = x^x(1 + lnx) #