Question

Помогите пожалуйста!

Answer 1

как то так!!!!...............

Answer 2

$f(x) = sqrt{6x + 7}$
$f'(x)= frac{1}{2 sqrt{6x + 7} } cdot (6x + 7)'= \ = frac{3}{ sqrt{6x + 7} } \ f'(5)= frac{3}{ sqrt{6 cdot 3 + 7} } = frac{3}{5} = 0.6$

$f(x)= cos ^{4} x \ f'(x)=4 : cos ^{3} x( cosx)' = \ = 4cos ^{3} x( - sin x) = \ = - 2sin2x : cos ^{2} x \ f'( frac{pi}{4} ) = - 2sin( frac{pi}{2}) : cos ^{2} ( frac{pi}{4} ) = \ = - 2 cdot 1 times {( frac{1}{ sqrt{2} }) }^{2} = - 1$