Question

помогите срочно нужно !!!​

Answer 1

1) f(x) = - 2x

f'(x) = - 2

2) f(x) = 4x²

f'(x) = 2 × 4x = 8x

3)

$f(x) = - \frac{6}{x {}^{4} } = - 6 {x}^{ - 4} \\ f'(x) = - 4 \times ( - 6)x {}^{ -4 - 1} = 24 {x}^{ - 5} = \frac{24}{ {x}^{5} }$

4) f(x) = 5sinx

f'(x) = 5cosx

5) f(x) = 7cosx

f'(x) = 7 × (-sinx) = - 7sinx

6)

$f(x) = \frac{5 {x}^{3} }{ \sin(x) } \\ f'(x) = \frac{(5 {x}^{3}) '\sin(x) - ( \sin(x))' \times 5 {x}^{3} }{ \sin {}^{2} (x) } = \\ = \frac{15 {x}^{2} \sin(x) - 5 {x}^{3} \cos(x) }{ \sin {}^{2} (x) }$

7)

$f(x) = \frac{4 \cos(x) }{ {x}^{2} } \\ f'(x) = \frac{ {x}^{2} (4 \cos(x) ') -4 \cos(x) ( {x}^{2}) '}{( {x}^{2}) {}^{2} } = \\ = \frac{ - 4 {x}^{2} \sin(x) - 8x \cos(x) }{ {x}^{4} }$

8)

$f(x) = \sqrt{x} \times \tan(x) \\ f'(x) = ( \sqrt{x} ) '\tan(x) + ( \tan(x) ) '\sqrt{x} = \frac{ \tan(x) }{ 2\sqrt{x} } + \frac{ \sqrt{x} }{ \cos {}^{2} (x) }$