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)#

1 Answer
Steve M
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) #

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