:::section[Using the Exponential Model] Hyperbolic expressions become manageable when one chooses among parity, exponential definitions, identities, and derivative rules. The following problems illustrate that choice. ::: :::solved-problem[Exact and Approximate Values] Since $\cosh$ is even, $$\begin{align*} \cosh(-2)&=\cosh2=\frac{e^2+e^{-2}}{2}\approx3.7622. \end{align*}$$ For logarithmic arguments, $$\begin{align*} \tanh(\ln4) &=\frac{4-\frac14}{4+\frac14} =\frac15{17},\\ \sinh(\ln3) &=\frac{3-\frac13}{2} =\frac43. \end{align*}$$ Likewise, $$\begin{align*} \cosh(-\ln2)&=\frac54,\\ \sinh(-3\ln2)&=-\sinh(\ln8)=-\frac{63}{16}. \end{align*}$$ ::: :::solved-problem[An Exponential Limit] To evaluate $$\begin{equation*} \lim_{x\to\infty}\frac{\cosh x}{e^x}, \end{equation*}$$ substitute the exponential definition: $$\begin{align*} \frac{\cosh x}{e^x} &=\frac12\left(1+e^{-2x}\right). \end{align*}$$ Since $e^{-2x}\to0$, the limit is $1/2$. ::: :::solved-problem[Composite Derivatives] For $y=\tanh(x^3+2)$, $$\begin{equation*} \frac{dy}{dx}=3x^2\operatorname{sech}^2(x^3+2). \end{equation*}$$ For $y=\ln(\cosh x)$, the chain rule gives $$\begin{equation*} \frac{dy}{dx}=\frac{\sinh x}{\cosh x}=\tanh x. \end{equation*}$$ ::: :::theorem[A Logarithmic Hyperbolic Identity] For every $x>0$, $$\begin{equation*} \tanh\left(\frac12\ln x\right)=\frac{x-1}{x+1}. \end{equation*}$$ ::: :::proof[Proof] Let $u=\frac12\ln x$. Then $e^{2u}=x$, and $$\begin{align*} \tanh u &=\frac{e^{2u}-1}{e^{2u}+1} =\frac{x-1}{x+1}. \end{align*}$$ The condition $x>0$ is required for $\ln x$. $\square$ ::: :::mistake[Exact Values Before Decimals] When an argument is a logarithm of a rational number, first use $e^{\ln a}=a$. This often produces an exact rational answer; a decimal should be secondary. ::: :::exercise[Practice Questions] 1. Evaluate $\cosh(\ln3)$ and $\tanh(\ln3)$ exactly. 2. Find $\displaystyle\lim_{x\to-\infty}e^x\cosh x$. 3. Differentiate $y=\ln(\operatorname{sech}x)$. 4. Solve $\tanh\left(\frac12\ln x\right)=\frac13$ for $x>0$. ::: :::answer[Answers and Guidance] 1. $$\begin{equation*} \cosh(\ln3)=\frac53,\qquad \tanh(\ln3)=\frac45. \end{equation*}$$ 2. $$\begin{equation*} e^x\cosh x=\frac12(e^{2x}+1)\longrightarrow\frac12. \end{equation*}$$ 3. $$\begin{equation*} y'=\frac{-\operatorname{sech}x\tanh x}{\operatorname{sech}x}=-\tanh x. \end{equation*}$$ 4. Using $(x-1)/(x+1)=1/3$ gives $3x-3=x+1$, hence $x=2$. ::: :::faq[Frequently Asked Questions] Q: Why are logarithmic arguments useful in hyperbolic functions? A: The exponential definitions convert $e^{\ln a}$ and $e^{-\ln a}$ into $a$ and $1/a$, often giving exact rational values. Q: Is $\ln(\cosh x)$ defined for every real $x$? A: Yes. Since $\cosh x\ge1$, its logarithm is defined for all real $x$. Q: Does $\cosh x/e^x$ have the same limit at both infinities? A: No. It tends to $1/2$ as $x\to\infty$, while it grows without bound as $x\to-\infty$. ::: :::call-to-action[Continue Learning] subtitle: Construct inverse hyperbolic functions and derive their logarithmic forms with precise real domains. button: Derive Inverse Logarithmic Forms|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/logarithmic-forms-of-inverse-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::