Question

ПРОШУ ПОМОГИТЕ..СРОЧНО!!

Answer 1

$f'(x)=( frac{x^2}{0,5^{1-2x}})'= (2^{1-2x}*x^2})'=(2^{1-2x})'*x^2+2^{1-2x}*(x^2)'= \ =-2^{1-2x}*ln2*2+2x*2^{1-2x}= 2^{2-2x}x-2^{2-2x}ln2= \ =4^{1-x}x-4^{1-x}ln2=4^{1-x}(x-ln2). \ f'(1)=4^{1-1}(1-ln2)=4^0(1-ln2)=1-ln2.$