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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