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

Observe that $$(a+b)^2-(a-b)^2=4ab$$
Set $a=e^x,b=e^{-x}$  and $\displaystyle \operatorname{sech}x=\frac1{\cosh x}=\frac2{e^x+e^{-x}}$
