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  "body": "\n## Proof\n\nWe differentiate the defining formulas with the standard rules.\n\n1. Write $e^t = \\exp(t)$ for the [[exponential function|0RMVS7RXZX]]. Let $E(x) = e^x$ and $F(x) = e^{-x}$ for real $x$.\n2. By [[$\\exp' = \\exp$|0EX9GFJ0B3]], $E$ is differentiable on $\\R$ with $E' = E$.\n3. The map $x \\mapsto -x$ is differentiable with derivative $-1$. By the [[chain rule|0RXX2PT1MQ]] and step 2, $F$ is differentiable on $\\R$ with $F'(x) = e^{-x} \\cdot (-1) = -F(x)$.\n4. By the definitions of the [[hyperbolic sine|0GWD21EWRW]] and the [[hyperbolic cosine|0EQFWKJ37J]], $\\sinh = \\frac{1}{2} E - \\frac{1}{2} F$ and $\\cosh = \\frac{1}{2} E + \\frac{1}{2} F$.\n5. By [[linearity of differentiation|0M0TKFRX2A]] and steps 2 and 3, $\\sinh$ is differentiable on $\\R$ with $\\sinh'(x) = \\frac{1}{2} e^x + \\frac{1}{2} e^{-x} = \\cosh x$.\n6. In the same way, $\\cosh$ is differentiable on $\\R$ with $\\cosh'(x) = \\frac{1}{2} e^x - \\frac{1}{2} e^{-x} = \\sinh x$.\n7. For each real $x$, $e^x \u003e 0$ and $e^{-x} \u003e 0$, since [[the exponential function is positive|0P9BNBW2W3]]. So $\\cosh x \u003e 0$, and in particular $\\cosh x \\ne 0$.\n8. By the definition of the [[hyperbolic tangent|0MWYQTZ07Z]], $\\tanh = \\sinh / \\cosh$.\n9. By the [[quotient rule|0BS0NJPKVE]] and steps 5, 6 and 7, $\\tanh$ is differentiable at each real $x$, with\n$$\\tanh'(x) = \\frac{\\cosh x \\cosh x - \\sinh x \\sinh x}{\\cosh^2 x} = \\frac{\\cosh^2 x - \\sinh^2 x}{\\cosh^2 x}.$$\n10. By the identity [[$\\cosh^2 - \\sinh^2 = 1$|06QCTRFBM0]], the numerator in step 9 is $1$. So $\\tanh'(x) = \\frac{1}{\\cosh^2 x}$.\n11. Dividing that identity by $\\cosh^2 x \\ne 0$ gives $1 - \\frac{\\sinh^2 x}{\\cosh^2 x} = \\frac{1}{\\cosh^2 x}$. So $\\frac{1}{\\cosh^2 x} = 1 - \\tanh^2 x$.\n12. Steps 5, 6, 10 and 11 give all three formulas, for every real $x$.\n\n## Idea\n\nThe exponential is its own derivative, and $e^{-x}$ has derivative $-e^{-x}$. So differentiating swaps the sign between the two terms, which turns $\\sinh$ into $\\cosh$ and back. For $\\tanh$, the quotient rule produces $\\cosh^2 - \\sinh^2$, which is $1$.\n",
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