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)