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，$y' = e^{-3x^2} \cdot (-6x) = -6x e^{-3x^2}$.
(4) $y=\ln (1+{x}^{2})$;
使用链式法则，$y' = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}$.
(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，$y' = \frac{1}{2} (a^2-x^2)^{-\frac{1}{2}} \cdot (-2x) = -\frac{x}{\sqrt{a^2-x^2}}$.
(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，$y' = \frac{1}{1+e^{2x}} \cdot e^x = \frac{e^x}{1+e^{2x}}$.
(9) $y={(\arcsin x)}^{2}$;
使用链式法则，$y' = 2\arcsin x \cdot \frac{1}{\sqrt{1-x^2}} = \frac{2\arcsin x}{\sqrt{1-x^2}}$.
(10) $y=\ln \cos x$;
使用链式法则，$y' = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.