$$\frac{d}{dx}(\operatorname{sech} x) = \frac{d}{dx} \left[ \frac{1}{\cosh x} \right] = \frac{-\sinh x}{\cosh^2 x} = -\operatorname{sech} x \tanh x \quad (8.24d)$$

$$\frac{d}{dx}(\operatorname{cosech} x) = \frac{d}{dx} \left[ \frac{1}{\sinh x} \right] = -\operatorname{cosech} x \coth x \quad (8.24e)$$

$$\frac{d}{dx}(\operatorname{coth} x) = \frac{d}{dx} \left[ \frac{\cosh x}{\sinh x} \right] = -\operatorname{cosech}^2 x \quad (8.24f)$$

按照与反三角函数相同的方法（见第 8.3.9 节），可以很容易地得到反双曲函数的下列导数。

$$\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1+x^2}} \quad (8.25a)$$

$$\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2-1}}, \quad x > 1 \quad (8.25b)$$

$$\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1-x^2}, \quad |x| < 1 \quad (8.25c)$$

例 8.20

求 $\frac{dy}{dx}$ 当 $y$ 由下式给出时

$$(a) \tanh 2x \quad (b) \cosh^2 x \quad (c) e^{-3x} \sinh 3x \quad (d) \sinh^{-1} \left[ \frac{3x}{4} \right]$$

解 (a) 引入中间变量 $z = 2x$ 得

$$y = \tanh z \quad \text{and} \quad z = 2x$$

$$\frac{dy}{dz} = \operatorname{sech}^2 z \quad \text{and} \quad \frac{dz}{dx} = 2$$

因此由链式法则

$$\frac{dy}{dx} = 2 \operatorname{sech}^2(2x)$$

(b) 引入中间变量 $z = \cosh x$ 得

$$y = z^2 \quad \text{and} \quad z = \cosh x$$

$$\frac{dy}{dz} = 2z \quad \text{and} \quad \frac{dz}{dx} = \sinh x$$

因此由链式法则

$$\frac{dy}{dx} = 2 \cosh x \sinh x = \sinh 2x$$