Hyperbolic functions

Explanation

• The hyperbolic functions are $\sinh x=\dfrac{e^x-e^{-x}}2$, $\cosh x=\dfrac{e^x+e^{-x}}2$ and $\tanh x=\dfrac{\sinh x}{\cosh x}$. Each has domain $\mathbb R$.
• The ranges are respectively $\mathbb R$, $[1,\infty)$ and $(-1,1)$.
• Both $\sinh x$ and $\tanh x$ are odd, pass through the origin and increase; $\cosh x$ is even, with minimum $(0,1)$.
• The graph of $\tanh x$ approaches the horizontal asymptotes $y=\pm1$ but never reaches them.
• Examiners expect sketches to show symmetry, intercepts, turning points and asymptotes, and exponential definitions may be used to calculate exact values or prove identities.

Explanation

• The fundamental derivatives are $\dfrac{d}{dx}(\sinh x)=\cosh x$, $\dfrac{d}{dx}(\cosh x)=\sinh x$ and $\dfrac{d}{dx}(\tanh x)=\operatorname{sech}^2x$.
• Chain, product and quotient rules apply in the usual way; unlike circular cosine, differentiating $\cosh x$ introduces no minus sign.
• Reverse the derivative rules for integration and divide by a constant inner derivative, so $\int\sinh(ax+b)\,dx=\dfrac1a\cosh(ax+b)+C$.
• Expressions may mix powers, products and radicals, as in the specification guidance, so algebraic simplification can be needed before choosing a rule.
• Examiners expect every chain-rule factor and the arbitrary constant in an indefinite integral.

Explanation

• An inverse hyperbolic function reverses a suitable one-to-one hyperbolic function. Thus $y=\operatorname{arsinh}x$ means $x=\sinh y$ and has domain and range $\mathbb R$.
• The function $\cosh x$ must first be restricted to $x\geq0$; consequently $\operatorname{arcosh}x$ has domain $[1,\infty)$ and range $[0,\infty)$.
• Finally, $y=\operatorname{artanh}x$ means $x=\tanh y$, with domain $(-1,1)$ and range $\mathbb R$.
• Applying a function and its inverse is valid only on these domains.
• Examiners expect domain checks when solving equations and the principal, non-negative value from $\operatorname{arcosh}$, not both signs from the even function $\cosh$.

Explanation

• The logarithmic forms are $\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$, $\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})$ for $x\geq1$, and $\operatorname{artanh}x=\dfrac12\ln\!\left(\dfrac{1+x}{1-x}\right)$ for $|x|<1$.
• They follow by replacing the hyperbolic function by exponentials, setting $u=e^y>0$, and solving the resulting quadratic for $u$.
• The positive exponential condition rejects the negative root.
• These forms turn inverse-hyperbolic values into exact logarithms and can prove identities, provided every square root and logarithm is properly defined.
• Examiners expect the domain restriction to accompany a derived form and any simplification of $\ln(z^2)$ to justify the sign of $z$.

Explanation

• Hyperbolic substitutions exploit $\cosh^2u-\sinh^2u=1$. For $\sqrt{x^2+a^2}$ with $a>0$, take $x=a\sinh u$, giving $dx=a\cosh u\,du$ and $\sqrt{x^2+a^2}=a\cosh u$.
• For $\sqrt{x^2-a^2}$ on the branch $x\geq a$, take $x=a\cosh u$, giving $dx=a\sinh u\,du$ and $\sqrt{x^2-a^2}=a\sinh u$.
• Hence the standard reciprocal-root integrals produce $\operatorname{arsinh}(x/a)+C$ and $\operatorname{arcosh}(x/a)+C$ on the appropriate domains.
• Associated integrals may require simplifying powers of $\sinh u$ and $\cosh u$ before reversing the substitution into an exact expression in $x$.
• Examiners expect $dx$, the radical and definite limits all to be transformed before integration.