Question

Помогите пожалуйста решить : f'(u)=((2u-u^1/2)^2)' даю 30 БАЛЛОВ!!!!!!!!!!!!!!!

Answer 1

$f'(u)=((2u-u^{\frac{1}{2}})^{2} )'=2(2u-u^{\frac{1}{2}})*(2u-u^{\frac{1}{2}})'=2(2u-u^{0,5})*(2-0,5u^{-0,5})=(2u-u^{0,5})(4-u^{-0,5})\\\\Otvet:\boxed{(2u-u^{0,5})(4-u^{-0,5})}$

Answer 2

f(g(x))' = f'(g(x)) * g'(x)

f(x) = x^n  f'(x) = n*x^(n-1)

(√x)' = (x^1/2)' = 1/2*x^-1/2 = 1/2√x

((2u -  √u )^2)' = 2(2u - √u)*(2u - √u)' = 2(2u - √u)*(2 - 1/(2√u)) = (2u - √u)*(2 - 1/√u)