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?

No. The answer I gave you before is right.

 

(ln(x) +1)x^x

Why is my answer wrong

Nevermind, our answers are the same.

 

y = x^x

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

 

 

and why would you use implicit differentiation?

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 :/

Negative infinity is NOT a real number.