:::section[Derivative Structure] The derivatives of $\sinh$ and $\cosh$ interchange the two functions. The remaining formulas follow from quotient and reciprocal rules together with $\cosh^2x-\sinh^2x=1$. ::: :::theorem[Derivatives of the Six Hyperbolic Functions] On their real domains, $$\begin{align*} \frac{d}{dx}\sinh x&=\cosh x, & \frac{d}{dx}\cosh x&=\sinh x,\\ \frac{d}{dx}\tanh x&=\operatorname{sech}^2x, & \frac{d}{dx}\coth x&=-\operatorname{csch}^2x,\\ \frac{d}{dx}\operatorname{sech}x&=-\operatorname{sech}x\tanh x, & \frac{d}{dx}\operatorname{csch}x&=-\operatorname{csch}x\coth x. \end{align*}$$ The formulas for $\coth$ and $\operatorname{csch}$ require $x\ne0$. ::: :::proof[Proof] From the exponential definitions, $$\begin{align*} \frac{d}{dx}\sinh x &=\frac12\left(e^x+e^{-x}\right)=\cosh x,\\ \frac{d}{dx}\cosh x &=\frac12\left(e^x-e^{-x}\right)=\sinh x. \end{align*}$$ Using the quotient rule, $$\begin{align*} \frac{d}{dx}\tanh x &=\frac{\cosh^2x-\sinh^2x}{\cosh^2x} =\operatorname{sech}^2x,\\ \frac{d}{dx}\coth x &=\frac{\sinh^2x-\cosh^2x}{\sinh^2x} =-\operatorname{csch}^2x. \end{align*}$$ Using the reciprocal rule, $$\begin{align*} \frac{d}{dx}\operatorname{sech}x &=-\frac{\sinh x}{\cosh^2x} =-\operatorname{sech}x\tanh x,\\ \frac{d}{dx}\operatorname{csch}x &=-\frac{\cosh x}{\sinh^2x} =-\operatorname{csch}x\coth x. \end{align*}$$ Hence all six formulas are established. $\square$ ::: :::solved-problem[Differentiate a Logarithmic Composition] To determine the derivative of $$\begin{equation*} y=\coth(\ln x),\qquad x>0, \end{equation*}$$ apply the chain rule: $$\begin{align*} \frac{dy}{dx} &=-\operatorname{csch}^2(\ln x)\frac{d}{dx}(\ln x)\\ &=-\frac{\operatorname{csch}^2(\ln x)}{x}. \end{align*}$$ ::: :::solved-problem[Product and Chain Rules Together] Let $$\begin{equation*} y=x^3\tanh^2(\sqrt{x}),\qquad x>0. \end{equation*}$$ Then $$\begin{align*} \frac{dy}{dx} &=3x^2\tanh^2(\sqrt{x}) +x^3\cdot2\tanh(\sqrt{x})\operatorname{sech}^2(\sqrt{x}) \frac{1}{2\sqrt{x}}\\ &=3x^2\tanh^2(\sqrt{x}) +x^{5/2}\tanh(\sqrt{x})\operatorname{sech}^2(\sqrt{x}). \end{align*}$$ ::: :::mistake[Differentiate the Inner Function] For $y=\tanh(g(x))$, the derivative is not merely $\operatorname{sech}^2(g(x))$. It is $$\begin{equation*} y'=\operatorname{sech}^2(g(x))g'(x). \end{equation*}$$ ::: :::exercise[Practice Questions] Differentiate each function on its real domain. 1. $y=\sinh(3x-1)$. 2. $y=\operatorname{sech}(x^2)$. 3. $y=x\cosh x-\sinh x$. 4. $y=\ln(\sinh x)$ for $x>0$. ::: :::answer[Answers and Guidance] 1. $$\begin{equation*} y'=3\cosh(3x-1). \end{equation*}$$ 2. $$\begin{equation*} y'=-2x\operatorname{sech}(x^2)\tanh(x^2). \end{equation*}$$ 3. Product differentiation gives $y'=\cosh x+x\sinh x-\cosh x=x\sinh x$. 4. $$\begin{equation*} y'=\frac{\cosh x}{\sinh x}=\coth x. \end{equation*}$$ ::: :::faq[Frequently Asked Questions] Q: Why does the derivative of $\cosh x$ not have a minus sign? A: Differentiating $e^{-x}$ introduces a minus sign, which cancels the plus sign pattern in the definition and produces $\sinh x$. Q: Where is the derivative of $\coth x$ valid? A: It is valid on each interval $(-\infty,0)$ and $(0,\infty)$, where $\coth x$ is defined. Q: Which identity is used in the derivative of $\tanh x$? A: The numerator from the quotient rule is $\cosh^2x-\sinh^2x=1$. ::: :::call-to-action[Continue Learning] subtitle: Apply the derivative table to exact values, limits, logarithmic compositions, and equations. button: Solve Differentiation Applications|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/applications-of-hyperbolic-differentiation button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::