微积分学/导数/导数练习

求导数

$y = \frac{(2 + 5 x)^{10}}{x}$

$\begin{matrix} y^{'} & = {[\frac{(2 + 5 x)^{10}}{x}]}^{'} \\ = - \frac{1}{x^{2}} (2 + 5 x)^{10} + \frac{1}{x} \cdot 10 (2 + 5 x)^{9} \cdot 5 \\ = - \frac{(2 + 5 x)^{10}}{x^{2}} + \frac{50 (2 + 5 x)^{9}}{x} \end{matrix}$

$y = x \cos (2 x + \frac{π}{2}) \sin (2 x + \frac{π}{2})$

$y = x (- \sin 2 x) \cos 2 x = - \frac{1}{2} x \sin 4 x$

$\begin{matrix} y^{'} & = {(- \frac{1}{2} x \sin 4 x)}^{'} \\ = - \frac{1}{2} \sin 4 x - \frac{1}{2} x \cos 4 x \cdot 4 \\ = - \frac{1}{2} \sin 4 x - 2 x \cos 4 x \end{matrix}$