Question

Найти производную функции
(4e^1-2x + sin3x/2 - x/2) ' = ?
(ln(3x-1)/2 - 3/√x^3)' = ?

Answer 1

Ответ:

1) $\boxed{ \displaystyle \boldsymbol{ \bigg( 4e^{1 - 2x} + \sin \frac{3x}{2} - \frac{x}{2} \bigg)' =-8e^{1 - 2x} + \frac{3\cos \bigg( \dfrac{3x}{2} \bigg)- 1}{2}} }$

2) $\boxed{ \displaystyle \boldsymbol{ \bigg (\frac{\ln(3x - 1)}{2} - \frac{3}{\sqrt{x^{3}} } \bigg)' = \frac{3}{6x - 2}+4,5x^{-2,5}}}$

Примечание:

Правила дифференцирования:

$(f_{1} \pm f_{2} \pm \ldots \pm f_{n})' = f_{1}' \pm f_{2}' \pm \ldots \pm f_{n}'$

$(kf)' = kf'$

$(f(g))' = g' f'(g)$

Таблица производных:

$\boxed{(x^{n})' = nx^{n -1}}$

$\boxed{(e^{x})' = e^{x}}$

$\boxed{(\ln x)' = \dfrac{1}{x} }$

$\boxed{(\sin x)' = \cos x}$

Пошаговое объяснение:

1) $\displaystyle \bigg( 4e^{1 - 2x} + \sin \frac{3x}{2} - \frac{x}{2} \bigg)' = \bigg( 4e^{1 - 2x} \bigg)' + \bigg( \sin \frac{3x}{2} \bigg)' - \bigg( \frac{x}{2} \bigg)' =$

$\displaystyle = 4 \bigg( e^{1 - 2x} \bigg)' + \bigg(\frac{3x}{2} \bigg)' \cos \frac{3x}{2}- \frac{1}{2} \bigg( x \bigg)' =$

$\displaystyle = 4 \bigg(1 - 2x \bigg)' e^{1 - 2x} + \frac{3}{2}\bigg(x \bigg)' \cos \frac{3x}{2}- \frac{1}{2} \cdot 1=$

$\displaystyle = 4\bigg( -2 \bigg) e^{1 - 2x} + \frac{3}{2} \cdot 1 \cdot \cos \frac{3x}{2} - \frac{1}{2} =-8e^{1 - 2x} + \frac{3\cos \bigg( \dfrac{3x}{2} \bigg)}{2} - \frac{1}{2} =$

$\displaystyle = -8e^{1 - 2x} + \frac{3\cos \bigg( \dfrac{3x}{2} \bigg)- 1}{2}$

2) $\displaystyle \bigg (\frac{\ln(3x - 1)}{2} - \frac{3}{\sqrt{x^{3}} } \bigg)' = \bigg(\frac{\ln(3x - 1)}{2} \bigg)' - \bigg(\frac{3}{\sqrt{x^{3}} } \bigg)' = \frac{1}{2} \bigg(\ln(3x - 1) \bigg)' - 3\bigg(\frac{1}{\sqrt{x^{3}} } \bigg)' =$

$\displaystyle = \frac{(3x - 1)'}{2(3x - 1)} - 3 \bigg( x^{-1,5} \bigg)' = \frac{3}{6x - 2} -3 \cdot (-1,5) \cdot x^{(-1,5 - 1)} = \frac{3}{6x - 2}+4,5x^{-2,5}$