Question

Найти значение производной функции f(x) в точке x0

Answer 1

Ответ:

$20)f'(x) = - \sin(3x - \frac{\pi}{2} ) \times 3 = - 3 \sin(3x - \frac{\pi}{2} )$

$f'(\frac{\pi}{3} ) = - 3 \sin(\pi - \frac{\pi}{2} ) = - 3 \sin( \frac{\pi}{2} ) = - 3$

$21)f'(x) = - {e}^{3 - x} + \frac{1}{ ln(2) \times (2x - 3)} \times 2$

$f'(2) = - {e}^{3 - 2} + \frac{1}{ ln(2) \times 1 } \times 2 = - e + \frac{2}{ ln(2) }$

$22)f'(x) = 3 {e}^{3x} (3 - 2x) - 2 {e}^{3x} = {e}^{3x} (9 - 6x - 2) = {e}^{3x} (7 - 6x)$

$f'(0) = {e}^{0} \times 7 = 7$

Answer 2

Ответ:

20. $-3$

21. $\frac{2}{ln2} -e$

22. $7$

Пошаговое объяснение:

20. $f(x)=cos(-\frac{\pi }{2} +3x)=cos(\frac\pi {2}-3x)=sin3x$

f'(x)=(sin3x)'=3cos(3x)

$f'(\frac{\pi}{3} )=3cos(\frac{3\pi }{3} )=3cos\pi=-3$

21. $f'(x)=(e^{3-x})'+(log_{2} (2x-3))'= -e^{3-x} +\frac{2}{(2x-3)ln2}$

$f'(2)=-e^{3-2}+\frac{2}{(2*2-3)ln2}=-e+\frac{2}{ln2} = \frac{2}{ln2} -e$

22. $f'(x)=(e^{3x} )'(3-2x)+e^{3x} (3-2x)'=3e^{3x} (3-2x)+e^{3x} (-2)=$

$=e^{3x} (9-6x-2)=e^{3x} (7-6x)$

$f'(0)=e^{3*0} (7-6*0)=e^{0} (7-0)=1*7=7$