blike Posted March 7, 2003 Posted March 7, 2003 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?
fafalone Posted March 7, 2003 Posted March 7, 2003 No. The answer I gave you before is right. (ln(x) +1)x^x
fafalone Posted March 7, 2003 Posted March 7, 2003 Because ln(x^x) <> x ln x for all real numbers (0 is the exception) and why would you use implicit differentiation?
blike Posted March 7, 2003 Author Posted March 7, 2003 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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