:::section[Inverse Derivatives] If $y=f^{-1}(x)$, then $x=f(y)$ and $$\begin{equation*} \frac{dy}{dx}=\frac{1}{f'(y)} \end{equation*}$$ whenever $f'(y)\ne0$. Hyperbolic identities then rewrite the result in terms of $x$. ::: :::theorem[Derivative Table] On the indicated intervals, $$\begin{align*} \frac{d}{dx}\sinh^{-1}x&=\frac{1}{\sqrt{1+x^2}}, &&x\in\mathbb{R},\\ \frac{d}{dx}\cosh^{-1}x&=\frac{1}{\sqrt{x^2-1}}, &&x>1,\\ \frac{d}{dx}\tanh^{-1}x&=\frac{1}{1-x^2}, &&|x|<1,\\ \frac{d}{dx}\coth^{-1}x&=\frac{1}{1-x^2}, &&|x|>1,\\ \frac{d}{dx}\operatorname{sech}^{-1}x&=-\frac{1}{x\sqrt{1-x^2}}, &&0<x<1,\\ \frac{d}{dx}\operatorname{csch}^{-1}x&=-\frac{1}{|x|\sqrt{1+x^2}}, &&x\ne0. \end{align*}$$ ::: :::proof[Derivative of Inverse Hyperbolic Sine] Let $y=\sinh^{-1}x$. Then $x=\sinh y$, and implicit differentiation gives $$\begin{equation*} 1=\cosh y\frac{dy}{dx}. \end{equation*}$$ Since $\cosh y>0$ and $\cosh^2y=1+\sinh^2y=1+x^2$, $$\begin{equation*} \frac{dy}{dx}=\frac1{\sqrt{1+x^2}}. \end{equation*}$$ $\square$ ::: :::proof[Derivative of Inverse Hyperbolic Cosine] Let $y=\cosh^{-1}x$, so $y\ge0$ and $x=\cosh y$. Then $$\begin{equation*} 1=\sinh y\frac{dy}{dx}. \end{equation*}$$ For $x>1$, one has $y>0$ and $\sinh y=\sqrt{\cosh^2y-1}=\sqrt{x^2-1}$. Hence $$\begin{equation*} \frac{dy}{dx}=\frac1{\sqrt{x^2-1}}. \end{equation*}$$ At $x=1$, the denominator vanishes, so no finite derivative exists. $\square$ ::: :::proof[Why an Absolute Value Appears for Inverse Cosecant] Let $y=\operatorname{csch}^{-1}x$. Then $x=\operatorname{csch}y$, and $$\begin{equation*} 1=-\operatorname{csch}y\coth y\frac{dy}{dx}. \end{equation*}$$ Thus $$\begin{equation*} \frac{dy}{dx}=-\frac{1}{x\coth y}. \end{equation*}$$ Since $\coth^2y=1+\operatorname{csch}^2y=1+x^2$ and $\coth y$ has the same sign as $x$, $$\begin{equation*} x\coth y=|x|\sqrt{1+x^2}. \end{equation*}$$ This gives the stated formula. $\square$ ::: :::mistake[Endpoints Belong to a Domain but Not Necessarily to a Derivative Interval] The function $\cosh^{-1}x$ is defined at $x=1$, and $\operatorname{sech}^{-1}x$ is defined at $x=1$. Their derivative formulas require open intervals because the radical denominator vanishes at the endpoint. ::: :::exercise[Practice Questions] Differentiate on the largest real interval where the expression is differentiable. 1. $y=\sinh^{-1}(2x)$. 2. $y=\cosh^{-1}(x^2)$. 3. $y=\tanh^{-1}(3x)$. 4. $y=\operatorname{csch}^{-1}(x^2+1)$. ::: :::answer[Answers and Guidance] 1. $$\begin{equation*} y'=\frac{2}{\sqrt{1+4x^2}}. \end{equation*}$$ 2. The real domain is $|x|\ge1$, and differentiability requires $|x|>1$: $$\begin{equation*} y'=\frac{2x}{\sqrt{x^4-1}}. \end{equation*}$$ 3. For $|x|<1/3$, $$\begin{equation*} y'=\frac{3}{1-9x^2}. \end{equation*}$$ 4. Since $x^2+1>0$, $$\begin{equation*} y'=-\frac{2x}{(x^2+1)\sqrt{1+(x^2+1)^2}}. \end{equation*}$$ ::: :::faq[Frequently Asked Questions] Q: Why is $(\cosh^{-1}x)'$ positive? A: The selected branch of $\cosh$ is increasing on $[0,\infty)$, so its inverse is increasing. Q: Why do $\tanh^{-1}$ and $\coth^{-1}$ have the same algebraic derivative? A: Both reduce to $1/(1-x^2)$, but they are defined on disjoint real domains. Q: Can the absolute value in $(\operatorname{csch}^{-1}x)'$ be omitted? A: No. Omitting it gives the wrong sign on the negative branch. ::: :::call-to-action[Continue Learning] subtitle: Place these derivative formulas inside composite functions and organize their complete domains and ranges. button: Apply Chain Rules and Ranges|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/chain-rules-and-ranges-for-inverse-hyperbolic-functions button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::