Question #8b077
1 Answer
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)#
