Question

найдите f'(1/2),если f(x)=(x^2+1)^2

Answer 1

f(1/2)=(1/2^2 +1)^2

f=(1/4+1)^2

f=1/16+1+1/2

f=(8+1+16)/16

f=25/16

Answer 2

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