Question

Найти производную срочноо

Answer 1

$1)f'(x) = \frac{1}{3} \times 6 {x}^{5} = 2 {x}^{5}$

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

$3)f'(x) = (6x + \frac{1}{3} {x}^{ - 1} ) '= 6 + \frac{1}{3} \times ( - {x}^{ - 2} ) = \\ = 6 - \frac{1}{3 {x}^{2} }$

$4)f'(x) = ( {x}^{2} (3 + x))' = (3 {x}^{2} + {x}^{3} )' = \\ = 6x + 3 {x}^{2}$

$5)f'(x) = 2(10 - 6x) \times (10 - 6x)' = \\ = 2(10 - 6x) \times ( - 6) = \\ = - 12(10 - 6 x) = 72x - 120$

Answer 2

Ответ:

)f

′

(x)=

3

1

×6x

5

=2x

5

\begin{gathered}2)f'(x) = - \frac{3}{ \sin {}^{2} (x) } \\ \end{gathered}

2)f

′

(x)=−

sin

2

(x)

3

\begin{gathered}3)f'(x) = (6x + \frac{1}{3} {x}^{ - 1} ) '= 6 + \frac{1}{3} \times ( - {x}^{ - 2} ) = \\ = 6 - \frac{1}{3 {x}^{2} } \end{gathered}

3)f

′

(x)=(6x+

3

1

x

−1

)

′

=6+

3

1

×(−x

−2

)=

=6−

3x

2

1

\begin{gathered}4)f'(x) = ( {x}^{2} (3 + x))' = (3 {x}^{2} + {x}^{3} )' = \\ = 6x + 3 {x}^{2} \end{gathered}

4)f

′

(x)=(x

2

(3+x))

′

=(3x

2

+x

3

)

′

=

=6x+3x

2

\begin{gathered}5)f'(x) = 2(10 - 6x) \times (10 - 6x)' = \\ = 2(10 - 6x) \times ( - 6) = \\ = - 12(10 - 6 x) = 72x - 120\end{gathered}

5)f

′

(x)=2(10−6x)×(10−6x)

′

=

=2(10−6x)×(−6)=

=−12(10−6x)=72x−120

Пошаговое объяснение:

Вот если не провильно то прости:(