TITLE: Hyperbolic Functions (derivative of $\tanh x$)
QUESTION [6 upvotes]: $$\sinh(x) = \frac{1}{2(e^x - e^{-x})}$$
$$\cosh(x) = \frac{1}{2(e^x + e^{-x}}$$
$$\tanh(x) = \frac{\sinh (x)}{\cosh (x)}$$
Prove:
$$\frac{d(\tanh(x))}{dx} = \frac{1}{(\cosh x)^2}$$ 
I got the derivative for $\tanh(x)$ as:
$$\left[ \frac{1}{2(e^x + e^{-x})}\right]^2 - \frac{{[ \frac{1}{2(e^x + e^{-x})}]^2}}{[ \frac{1}{2(e^x + e^{-x})}]^2}$$

REPLY [4 votes]: $$\dfrac{d(f/g)}{dx}=\dfrac{gf^\prime-fg^\prime}{g^2}$$
Set $f=\sinh,g=\cosh$ to get
$$\dfrac{d\tanh}{dx}=\dfrac{\cosh\cdot\sinh^\prime-\sinh\cdot\cosh^\prime}{\cosh^2}$$
Now,
$$\sinh^\prime=\dfrac{1}{2}(e^x+e^{-x})=\cosh\\
\cosh^\prime=\dfrac{1}{2}(e^x-e^{-x})=\sinh$$
Thus,
$$\dfrac{d\tanh}{dx}=\dfrac{\cosh^2-\sinh^2}{\cosh^2}=1-\left(\dfrac{\sinh}{\cosh}\right)^2=1-\tanh^2$$
Now, 
$$\dfrac{1}{\cosh^2}=1-\tanh^2$$
(Proof:
$$1-\dfrac{\sinh^2 x}{\cosh^2 x}= \dfrac{\cosh^2 x-\sinh^2 x}{\cosh^2 x}$$ 
Since $(\cosh^2 x) - (\sinh^2 x) = 1$,
$$\dfrac{1}{\cosh^2 x} = {\operatorname{sech}^2 x} $$)
Thus,
$$\boxed{\dfrac{d\tanh}{dx}=1-\tanh^2=\dfrac{1}{\cosh^2 x}}$$