Asked by Sean
Use logarithmic differentiation to find derivative of y=x^x^2 x > 0. Please show all steps so I can follow the steps. Thank you.!
Answers
Answered by
Steve
y = x^x^2
lny = x^2 ln x
1/y y' = 2x lnx + x^2/x
= 2x lnx + x
y' = xy(2lnx + 1)
y' = x x^x^2 (2lnx + 1)
lny = x^2 ln x
1/y y' = 2x lnx + x^2/x
= 2x lnx + x
y' = xy(2lnx + 1)
y' = x x^x^2 (2lnx + 1)
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