Question

Помогите пожалуйста!

Answer 1

как то так!!!!...............

Answer 2

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

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