Question

f(x) =e^2x *lnx^2. Найти f' (1)

Answer 1

$f(x)=e^{2x}*lnx^2\\f'(x)=(e^{2x})'*lnx^2+e^{2x}*(lnx^2)'=e^{2x}*(2x)'*lnx^2+e^{2x}*\frac{1}{x^2}*(x^2)'$
$=e^{2x}(2*lnx^2+\frac{2x}{x^2})=e^{2x}(lnx^4+\frac{2}{x})$
$f'(1)=e^{2*1}(ln1^4+\frac{2}{1})=e^2(0+2)=2e^2$