blike Posted March 7, 2003 Share 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? Link to comment Share on other sites More sharing options...
fafalone Posted March 7, 2003 Share Posted March 7, 2003 No. The answer I gave you before is right. (ln(x) +1)x^x Link to comment Share on other sites More sharing options...
blike Posted March 7, 2003 Author Share Posted March 7, 2003 Why is my answer wrong Link to comment Share on other sites More sharing options...
blike Posted March 7, 2003 Author Share Posted March 7, 2003 Nevermind, our answers are the same. y = x^x Link to comment Share on other sites More sharing options...
fafalone Posted March 7, 2003 Share 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? Link to comment Share on other sites More sharing options...
blike Posted March 7, 2003 Author Share 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 :/ Link to comment Share on other sites More sharing options...
fafalone Posted March 7, 2003 Share Posted March 7, 2003 Negative infinity is NOT a real number. Link to comment Share on other sites More sharing options...
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