Question #8b077

1 Answer
Gaurav
Oct 22, 2014

#y'=1/(2*sqrtx)*(ln(xe^x)+2+2x)#

Explanation :

#y=x^(1/2)*ln(xe^x)#

Using Product Rule, which is

#y=f(x)*g(x)#

differentiating with respect to #x#,

#y'=f'(x)*g(x)+f(x)*g'(x)#

Now similarly following for the given problem,

#y'=(x^(1/2))'ln(xe^x)+x^(1/2)(ln(xe^x))'#

#y'=1/2x^(-1/2)*ln(xe^x)+x^(1/2)*1/(xe^x)(xe^x)'#

#y'=1/2*1/x^(1/2)*ln(xe^x)+x^(1/2)*1/(xe^x)(e^x+xe^x)#

#y'=1/2*1/x^(1/2)*ln(xe^x)+1/(x^(1/2)e^x)(e^x+xe^x)#

#y'=1/2*1/x^(1/2)*ln(xe^x)+1/(x^(1/2))(1+x)#

#y'=1/2*1/x^(1/2)*(ln(xe^x)+2+2x)#

#y'=1/(2*sqrtx)*(ln(xe^x)+2+2x)#

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