求下列函数的导数： (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4; (1) =((2x+5))^4.

步骤 1：求导数
(1) $y={(2x+5)}^{4}$;
使用链式法则，$y' = 4{(2x+5)}^{3} \cdot 2 = 8{(2x+5)}^{3}$.
(2) $y=\cos (4-3x)$;
使用链式法则，$y' = -\sin (4-3x) \cdot (-3) = 3\sin (4-3x)$.
(3) $y={e}^{-3{x}^{2}}$;
使用链式法则，$y' = {e}^{-3{x}^{2}} \cdot (-6x) = -6x{e}^{-3{x}^{2}}$.
(4) $y=\ln (1+{x}^{2})$;
使用链式法则，$y' = \dfrac{1}{1+{x}^{2}} \cdot 2x = \dfrac{2x}{1+{x}^{2}}$.
(5) $y={\sin }^{2}x$;
使用链式法则，$y' = 2\sin x \cdot \cos x = \sin 2x$.
(6) $y=\sqrt {{a}^{2}-{x}^{2}}$;
使用链式法则，$y' = \dfrac{1}{2}({a}^{2}-{x}^{2})^{-\frac{1}{2}} \cdot (-2x) = -x({a}^{2}-{x}^{2})^{-\frac{1}{2}}$.
(7) $y=\tan {x}^{2}$;
使用链式法则，$y' = {\sec }^{2}{x}^{2} \cdot 2x = 2x{\sec }^{2}{x}^{2}$.
(8) $y=\arctan {e}^{x}$;
使用链式法则，$y' = \dfrac{1}{1+{e}^{2x}} \cdot {e}^{x} = \dfrac{{e}^{x}}{1+{e}^{2x}}$.
(9) $y={(\arcsin x)}^{2}$;
使用链式法则，$y' = 2\arcsin x \cdot \dfrac{1}{\sqrt{1-x^2}} = \dfrac{2\arcsin x}{\sqrt{1-x^2}}$.
(10) $y=\ln \cos x$;
使用链式法则，$y' = \dfrac{1}{\cos x} \cdot (-\sin x) = -\tan x$.