:::section[The Inner Function Controls the Domain] Let $t=t(x)$ be differentiable. A chain-rule formula is meaningful only where $t(x)$ belongs to the real domain of the outer inverse hyperbolic function. ::: :::theorem[Chain-Rule Formulas] Where the stated conditions hold, $$\begin{align*} \frac{d}{dx}\sinh^{-1}t &=\frac{t'}{\sqrt{1+t^2}},\\ \frac{d}{dx}\cosh^{-1}t &=\frac{t'}{\sqrt{t^2-1}}, &&t>1,\\ \frac{d}{dx}\tanh^{-1}t &=\frac{t'}{1-t^2}, &&|t|<1,\\ \frac{d}{dx}\coth^{-1}t&=\frac{t'}{1-t^2}, &&|t|>1,\\ \frac{d}{dx}\operatorname{sech}^{-1}t &=-\frac{t'}{t\sqrt{1-t^2}}, &&0<t<1,\\ \frac{d}{dx}\operatorname{csch}^{-1}t &=-\frac{t'}{|t|\sqrt{1+t^2}}, &&t\ne0. \end{align*}$$ ::: :::proof[Proof Pattern] For example, set $y=\sinh^{-1}(t(x))$. Then $$\begin{equation*} \frac{dy}{dt}=\frac{1}{\sqrt{1+t^2}}. \end{equation*}$$ By the chain rule, $$\begin{equation*} \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx} =\frac{t'}{\sqrt{1+t^2}}. \end{equation*}$$ The remaining formulas follow in the same way from the derivative table, with their domain restrictions retained. $\square$ ::: :::key-formula[Domains and Ranges of Principal Inverse Functions] $$\begin{align*} \operatorname{Dom}(\sinh^{-1})&=\mathbb{R}, &\operatorname{Ran}(\sinh^{-1})&=\mathbb{R},\\ \operatorname{Dom}(\cosh^{-1})&=[1,\infty), &\operatorname{Ran}(\cosh^{-1})&=[0,\infty),\\ \operatorname{Dom}(\tanh^{-1})&=(-1,1), &\operatorname{Ran}(\tanh^{-1})&=\mathbb{R},\\ \operatorname{Dom}(\coth^{-1})&=(-\infty,-1)\cup(1,\infty), &\operatorname{Ran}(\coth^{-1})&=\mathbb{R}\setminus\{0\},\\ \operatorname{Dom}(\operatorname{sech}^{-1})&=(0,1], &\operatorname{Ran}(\operatorname{sech}^{-1})&=[0,\infty),\\ \operatorname{Dom}(\operatorname{csch}^{-1})&=\mathbb{R}\setminus\{0\}, &\operatorname{Ran}(\operatorname{csch}^{-1})&=\mathbb{R}\setminus\{0\}. \end{align*}$$ ::: :::solved-problem[Differentiate with a Domain Check] Let $$\begin{equation*} y=\tanh^{-1}(2x-1). \end{equation*}$$ The real domain requires $|2x-1|<1$, hence $0<x<1$. On this interval, $$\begin{align*} y' &=\frac{2}{1-(2x-1)^2} =\frac{1}{2x(1-x)}. \end{align*}$$ ::: :::mistake[Do Not State Only an Algebraic Formula] A derivative such as $2/[1-(2x-1)^2]$ is incomplete without the interval $0<x<1$, because the outer inverse function is not real outside that interval. ::: :::exercise[Practice Questions] 1. Differentiate $y=\sinh^{-1}(x^3)$. 2. Differentiate $y=\cosh^{-1}(2x)$ and state the differentiability interval. 3. Differentiate $y=\operatorname{sech}^{-1}(e^{-x})$ on its real differentiability interval. 4. Determine the range of $\coth^{-1}x$. ::: :::answer[Answers and Guidance] 1. $$\begin{equation*} y'=\frac{3x^2}{\sqrt{1+x^6}}. \end{equation*}$$ 2. Since $2x>1$, the interval is $x>1/2$, and $$\begin{equation*} y'=\frac{2}{\sqrt{4x^2-1}}. \end{equation*}$$ 3. The condition $0<e^{-x}<1$ gives $x>0$. Then $$\begin{equation*} y'=\frac{1}{\sqrt{1-e^{-2x}}}. \end{equation*}$$ 4. The range is $(-\infty,0)\cup(0,\infty)$. ::: :::faq[Frequently Asked Questions] Q: Why is the endpoint excluded from the chain-rule formula for inverse secant? A: At the endpoint $t=1$, the factor $\sqrt{1-t^2}$ is zero, so the derivative is not finite. Q: Is the range of $\coth^{-1}$ all real numbers? A: No. The value zero is excluded because $\coth x$ is undefined at $x=0$ and never has an inverse output of zero. Q: Should domain restrictions be checked before or after differentiating? A: Before differentiating, so that the real-valued function and the interval of validity are known. ::: :::call-to-action[Continue Learning] subtitle: Combine domain checks with product, quotient, and chain rules in more demanding applications. button: Solve Inverse Differentiation Applications|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/applications-of-inverse-hyperbolic-differentiation button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::