Question

Найдите f''(x), если:​

Answer 1

1.

$f(x) = 5 {x}^{6} + 6 {x}^{2} + 4$

$f'(x) = 30 {x}^{7} + 12x + 0 = 30 {x}^{7} + 12x$

$f''(x) = 210 {x}^{6} + 12$

2.

$f(x) = \sin {}^{3} (x + 3)$

$f'(x) = 3 \sin {}^{2} (x + 3) \times ( \sin(x + 3))' = \\ = 3 \sin {}^{2} (x + 3) \cos(x + 3)$

$f''(x) = (3 \sin {}^{2} (x + 3) )'\times \cos(x + 3) + ( \cos {}^{} (x + 3) ) '\times 3 { \sin}^{2} ( x+ 3) = \\ = 6 \sin(x + 3) \times \cos(x + 3) \times \cos(x + 3) - \sin(x + 3) \times 3 \sin {}^{2} (x + 3) = \\ = 3 \sin(2x + 6) \cos(x + 6) - 3 \sin {}^{2} (x + 3)$

3.

$f(x) = {e}^{tgx}$

$f'(x) = {e}^{tgx} \times (tgx) ' = \frac{ {e}^{tgx} }{ \cos {}^{2} (x) }$

$f''(x) = \frac{( {e}^{tgx}) '\times \cos {}^{2} (x) - ( \cos {}^{2} (x))' e {}^{tgx} }{ \cos {}^{4} (x) } = \\ = \frac{ \frac{ {e}^{tgx} }{ \cos {}^{2} (x) } \times \cos {}^{2} (x) - 2 \cos(x)( - \sin(x)) \times {e}^{2tgx} }{ \cos {}^{4} (x) } = \\ = \frac{ {e}^{tgx} (1 + \sin(2x)) }{ \cos {}^{4} (x) }$