TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Addition formulas, identities, parity, and derivatives of the hyperbolic functions

Statement

For all real $x, y$ , $\sinh (x + y) = \sinh x \cosh y + \cosh x \sinh y, \cosh (x + y) = \cosh x \cosh y + \sinh x \sinh y,$ $\cosh^{2} x - \sinh^{2} x = 1, (\sinh x)^{'} = \cosh x, (\cosh x)^{'} = \sinh x .$ Moreover $\sinh : R \to R$ is odd, strictly increasing, and onto; $\cosh : R \to [1, \infty)$ is even and positive, and its restriction to $[0, \infty)$ is strictly increasing and onto $[1, \infty)$ ; and $\tanh : R \to (- 1, 1)$ is strictly increasing and onto. On their declared domains, $(\tanh x)^{'} = {sech}^{2} x, (\coth x)^{'} = - {csch}^{2} x, (sech x)^{'} = - sech x \tanh x, (csch x)^{'} = - csch x \coth x .$

Facts & Assumptions

Given: Real numbers $x, y$ .

[L1]

$\exp (u + v) = \exp u \exp v$ , $\exp (- u) = 1 / \exp u$ , and $\exp u > 0$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

[L2]

$\exp^{'} = \exp$ , and the chain, product, quotient, and algebra rules differentiate the displayed formulas (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L3]

If $a < b$ and $f : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then some $c \in (a, b)$ satisfies $f (b) - f (a) = f^{'} (c) (b - a)$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[L4]

$\exp x \to \infty$ as $x \to \infty$ and $\exp x \to 0$ as $x \to - \infty$ (The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ ).

[L5]

A continuous real function takes every value between two values on a closed interval (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[L6]

For every real $x$ , $\sinh x = (\exp x - \exp (- x)) / 2$ and $\cosh x = (\exp x + \exp (- x)) / 2$ ; $\tanh x = \sinh x / \cosh x$ and $sech x = 1 / \cosh x$ ; and, when $x \neq 0$ , $\coth x = \cosh x / \sinh x$ and $csch x = 1 / \sinh x$ . Moreover $\cosh x > 0$ for every $x$ , and $\sinh x \neq 0$ when $x \neq 0$ (The six hyperbolic functions and their natural domains).

[L7]

Differentiability implies continuity (A function differentiable at $c$ is continuous at $c$ ).

Proof

technique · direct

1.1

Substitute the exponential definitions of [L6] and use [L1]; collecting terms gives both addition formulas, parity, and $\cosh^{2} x - \sinh^{2} x = 1$ .

L1L6algebra

2.1

Differentiating the exponential definitions of [L6] gives $(\sinh x)^{'} = \cosh x$ and $(\cosh x)^{'} = \sinh x$ ; on the domains supplied by [L6], differentiating the quotients and using step 1.1 gives the four displayed reciprocal-function derivatives.

step 1.1L2L6algebra

2.2

The exponential formulas give $\sinh x \to \infty$ , $\cosh x \to \infty$ , and $\tanh x \to 1$ as $x \to \infty$ ; oddness gives the corresponding limits $- \infty$ and $- 1$ at $- \infty$ .

step 1.1L1L4algebra

3.1

By [L6], $\cosh x > 0$ and $\sinh x$ is nonzero away from $0$ ; the defining formula gives $\sinh 0 = 0$ . Thus $\sinh^{'} = \cosh > 0$ everywhere. For $a < b$ , step 2.1 and [L7] give the hypotheses of [L3] for $\sinh$ on $[a, b]$ , so for some $c \in (a, b)$ , $\sinh b - \sinh a = \cosh c (b - a) > 0$ . Hence $\sinh$ is strictly increasing.

step 2.1L3L6L7algebra

4.1

The oddness and strict increase of $\sinh$ make $\sinh x > 0$ for $x > 0$ . Thus for $0 \leq a < b$ , step 2.1 and [L7] let [L3] give $\cosh b - \cosh a = \sinh c (b - a) > 0$ for some $c \in (a, b)$ , so $\cosh$ is strictly increasing on $[0, \infty)$ . Also $sech x > 0$ , so for $a < b$ the same argument gives $\tanh b - \tanh a = {sech}^{2} c (b - a) > 0$ for some $c \in (a, b)$ ; hence $\tanh$ is strictly increasing.

step 1.1step 2.1step 3.1L1L3L6L7

5.1

The functions are continuous by step 2.1 and [L7]. Their monotonicity, the values $\cosh 0 = 1$ , and the endpoint limits of step 2.2 let the intermediate value theorem give exactly the three stated ranges.

step 2.1step 3.1step 4.1step 2.2L5L7∎

Depends on

Used by

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)