Question

на фото.....................................................................

Answer 1

По правилу дифференцирования 
$y=x^6*ln(ex)\\ y'=x^6'ln(ex)+x^6*ln(ex)'=6x^5*ln(ex)+x^5=x^5(6ln(ex)+1)\\ y'(e^2)=e^{10}*(6ln(e*e^2)+1)=e^{10}(6*3+1)=19*e^{10}$

Answer 2

$y=x^6ln(ex)$
$y'=(x^6ln(ex))'=(x^6)'*ln(ex)+x^6*(ln(ex))'=$
 $=6*x^5*ln(ex)+x^6*\frac{1}{x}$
$x_o=e^2$
$y'(e^2)=6*(e^2)^5*ln(e^3)+\frac{(e^2)^6}{e^2}=18e^{10}+e^{10}=19e^{10}$