:::section[Strategy for Applications] For each expression, first determine the real domain of the outer inverse function. Then identify the required product, quotient, or chain rules. Only after the domain has been fixed should the derivative be simplified. This order prevents a formal derivative from being mistaken for the derivative of a real-valued function. ::: :::solved-problem[Inverse Sine after Tangent] Let $$\begin{equation*} f(x)=\sinh^{-1}(\tan x). \end{equation*}$$ The tangent is defined when $x\ne \frac{\pi}{2}+k\pi$, where $k\in\mathbb{Z}$. Since $\sinh^{-1}$ accepts every real input, $$\begin{align*} f'(x) &=\frac{\sec^2 x}{\sqrt{1+\tan^2x}}\\ &=\frac{\sec^2x}{\sqrt{\sec^2x}}\\ &=\frac{\sec^2x}{|\sec x|}\\ &=|\sec x|. \end{align*}$$ The absolute value is necessary because the principal square root is nonnegative. ::: :::solved-problem[Inverse Cosine after Inverse Sine] Let $$\begin{equation*} y=\cosh^{-1}(\sinh^{-1}x). \end{equation*}$$ The outer inverse cosine requires $\sinh^{-1}x\ge 1$, so $x\ge \sinh 1$. Differentiability requires the strict inequality $x>\sinh 1$. Therefore, $$\begin{equation*} \frac{dy}{dx} = \frac{1} {\sqrt{(\sinh^{-1}x)^2-1}\sqrt{1+x^2}}, \qquad x>\sinh 1. \end{equation*}$$ ::: :::solved-problem[An Arctangent-Hyperbolic Composition] Let $$\begin{equation*} y=\tan^{-1}\left(\tanh\frac{x}{2}\right). \end{equation*}$$ Using the chain rule, $$\begin{align*} \frac{dy}{dx} &= \frac{\frac12\operatorname{sech}^2(x/2)} {1+\tanh^2(x/2)}\\ &= \frac{1}{2\cosh x}. \end{align*}$$ The final step follows from $$\begin{equation*} \frac{\operatorname{sech}^2u}{1+\tanh^2u} = \frac{1}{\cosh 2u}. \end{equation*}$$ ::: :::theorem[Useful Real Identities] With the stated domain restrictions, $$\begin{align*} \coth^{-1}\left(\frac{2}{x}\right) &= \sinh^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right), &&0<|x|<2,\\ \sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) &=\tanh^{-1}x, &&|x|<1,\\ \sinh^{-1}x &=\cosh^{-1}\sqrt{1+x^2}, &&x\ge 0. \end{align*}$$ For arbitrary real $x$, the last equality must be written $$\begin{equation*} |\sinh^{-1}x|=\cosh^{-1}\sqrt{1+x^2}. \end{equation*}$$ ::: :::proof[Proof of the Third Identity] Let $u=\sinh^{-1}x$. Then $\sinh u=x$ and $$\begin{equation*} \cosh u=\sqrt{1+\sinh^2u}=\sqrt{1+x^2}. \end{equation*}$$ If $x\ge0$, then $u\ge0$, which lies in the principal range of $\cosh^{-1}$. Hence $$\begin{equation*} u=\cosh^{-1}\sqrt{1+x^2}. \end{equation*}$$ Therefore the stated identity holds for $x\ge0$. For general real $x$, the principal inverse cosine returns $|u|$, which gives the absolute-value version. $\square$ ::: :::solved-problem[Solving a Linear Hyperbolic Equation] Given that $$\begin{equation*} 5\sinh x-\cosh x=5, \end{equation*}$$ write $s=\sinh x$ and $c=\cosh x$. Then $c=5s-5$. Substituting into $c^2-s^2=1$ gives $$\begin{align*} (5s-5)^2-s^2&=1,\\ 24s^2-50s + 24&=0,\\ (3s-4)(4s-3)&=0. \end{align*}$$ The value $s=3/4$ would give $c=-5/4$, which is impossible because $\cosh x>0$. Hence $$\begin{equation*} \sinh x=\frac43,\qquad \cosh x=\frac53,\qquad \tanh x=\frac45. \end{equation*}$$ ::: :::mistake[Formal Differentiation Does Not Create a Real Function] The expression $\tanh^{-1}(\coth x)$ is not real for any $x\ne0$, because $|\coth x|>1$ while the real domain of $\tanh^{-1}$ is $(-1,1)$. A formally simplified derivative therefore does not describe a real-valued composition. ::: :::exercise[Practice Questions] 1. Differentiate $y=\operatorname{csch}^{-1}(5\sqrt{x})$ for $x>0$. 2. Differentiate $y=\tanh^{-1}\left(\frac{x^2-1}{x^2+1}\right)$ for $x>0$. 3. Prove that $\coth^{-1}(\coth x)=x$ for $x\ne0$, and differentiate both sides. 4. Given that $\cosh x=\sec\theta$ and $\sinh x=\tan\theta$, prove that $\sec\theta+\tan\theta=e^x$. ::: :::answer[Answers and Guidance] 1. $$\begin{equation*} y'=-\frac{1}{2x\sqrt{1+25x}}. \end{equation*}$$ 2. Let $u=(x^2-1)/(x^2+1)$. Then $u'=4x/(x^2+1)^2$ and $1-u^2=4x^2/(x^2+1)^2$, so $$\begin{equation*} y'=\frac1x. \end{equation*}$$ 3. The principal inverse cotangent has range $\mathbb{R}\setminus\{0\}$, which contains each admissible $x$. Hence the composition law holds on both intervals $(-\infty,0)$ and $(0,\infty)$; differentiating gives $1=1$. 4. Substitute the two given relations into $\cosh x+\sinh x=e^x$. ::: :::faq[Frequently Asked Questions] Q: Why does $|\sec x|$ appear in the first derivative? A: Because $\sqrt{\sec^2x}=|\sec x|$, not $\sec x$ on intervals where secant is negative. Q: Can an inverse-hyperbolic identity be stated without a domain? A: No. Radicals, logarithms, and principal inverse branches can restrict an identity to a proper subset of the real line. Q: Why is the root giving $\cosh x<0$ rejected? A: The exponential definition gives $\cosh x=(e^x+e^{-x})/2>0$ for every real $x$. ::: :::call-to-action[Continue Learning] subtitle: Turn the differentiation table into a basic library of hyperbolic antiderivatives and verify each formula by differentiation. button: Learn Hyperbolic Integration Formulas|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/integration-formulas-for-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::