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Posted

question:

 

y= x^x

 

"solve using logarithmic differentiation"

 

ln y = ln (x^x)

 

ln y = x ln (x)

 

1/y dy/dx = (1)(ln(x)) + x(1/x)

 

1/y dy/dx = lnx + 1

 

dy/dx = y [ln (x) + 1]

 

is this right?

Posted
Because ln(x^x) <> x ln x for all real numbers (0 is the exception)

 

ln (x^x) = x ln x for all real numbers.

 

 

i don't know, because he said to :/

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