document.write( "Question 959170: Hello friends , Can someone help me with this problem? I understand almost all of the problem up to the last part . I am not sure how they are getting the simplification at the end . they ask me to differentiate y=x^lnx .
\n" ); document.write( "I understand enough to take the natural log of both sides lny=lmx^lnx
\n" ); document.write( "lny=lnxlnx =lny=(lnx)^2 then I take the derivative of both sides 1/ydy/dx=2ln(x)(1/x) = y'= 2ylnx(1/x)= I don't understand how they are simplifying it to
\n" ); document.write( "y'=2x^lnx-1lnx because I can not understand what happens to the derivative of lnx . it is 1/x their solution makes that 1/x vanish . I do not get the simplification part . I understand that they substituted the y for x^lnx but what happened to the derivative 1/x . I guessed that they used the power rule and raised it to the x^-1 power but how do you put it in the final simplification . I am confused . becaue x^-1 would be multiplied times x^lnx . How does that work ? \n" ); document.write( "

Algebra.Com's Answer #586289 by rothauserc(4718) $\"\"$ $\"About$

You can put this solution on YOUR website!
y=x^(ln(x))
\n" ); document.write( "ln(y)=ln(x^(ln(x))
\n" ); document.write( "ln(y)=ln(x)ln(x)
\n" ); document.write( "implicit differentiation+product rule yields
\n" ); document.write( "(1/y)y'=(1/x)ln(x)+(1/x)ln(x)
\n" ); document.write( "(1/y)y'=(2/x)ln(x)
\n" ); document.write( "y'=(2/x)yln(x)
\n" ); document.write( "=(2yln(x))/x
\n" ); document.write( " \n" ); document.write( "

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