{"id":494,"date":"2025-02-13T19:45:27","date_gmt":"2025-02-13T19:45:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-of-the-hyperbolic-functions-learn-it-2\/"},"modified":"2025-02-13T19:45:27","modified_gmt":"2025-02-13T19:45:27","slug":"calculus-of-the-hyperbolic-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-of-the-hyperbolic-functions-learn-it-2\/","title":{"raw":"Calculus of the Hyperbolic Functions: Learn It 2","rendered":"Calculus of the Hyperbolic Functions: Learn It 2"},"content":{"raw":"\n

Calculus of Inverse Hyperbolic Functions<\/h2>\n

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. The inverse hyperbolic functions have specific domain and range restrictions, summarized in the table below:<\/p>\n\n\n\n\n\n\n\n\n\n\n
Domains and Ranges of the Inverse Hyperbolic Functions<\/caption>\n
Function<\/th>\nDomain<\/th>\nRange<\/th>\n<\/tr>\n<\/thead>\n
[latex]{\\text{sinh}}^{-1}x[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{cosh}}^{-1}x[\/latex]<\/td>\n[latex](1,\\infty )[\/latex]<\/td>\n[latex][0,\\infty )[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{tanh}}^{-1}x[\/latex]<\/td>\n[latex](-1,1)[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{coth}}^{-1}x[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,-1)\\cup (1,\\infty )[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{sech}}^{-1}x[\/latex]<\/td>\n[latex](0\\text{, 1})[\/latex]<\/td>\n[latex][0,\\infty )[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{csch}}^{-1}x[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\n[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The graphs of the inverse hyperbolic functions are shown in the following figure.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"958\"]\"This Figure 2. Graphs of the inverse hyperbolic functions.[\/caption]\n\n

To find the derivatives of the inverse functions, we use implicit differentiation. We have<\/p>\n

[latex]\\begin{array}{ccc}\\hfill y& =\\hfill & {\\text{sinh}}^{-1}x\\hfill \\\\ \\hfill \\text{sinh}y& =\\hfill & x\\hfill \\\\ \\hfill \\frac{d}{dx}\\text{sinh}y& =\\hfill & \\frac{d}{dx}x\\hfill \\\\ \\hfill \\text{cosh}y\\frac{dy}{dx}& =\\hfill & 1.\\hfill \\end{array}[\/latex]<\/div>\n

Recall that [latex]{\\text{cosh}}^{2}y-{\\text{sinh}}^{2}y=1,[\/latex] so [latex]\\text{cosh}y=\\sqrt{1+{\\text{sinh}}^{2}y}.[\/latex] Then,<\/p>\n

[latex]\\frac{dy}{dx}=\\frac{1}{\\text{cosh}y}=\\frac{1}{\\sqrt{1+{\\text{sinh}}^{2}y}}=\\frac{1}{\\sqrt{1+{x}^{2}}}.[\/latex]<\/div>\n

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. <\/p>\n

\n

derivatives of inverse hyperbolic functions<\/h3>\n\n\n\n\n\n\n\n\n\n\n
Derivatives of the Inverse Hyperbolic Functions<\/caption>\n
[latex]f(x)[\/latex]<\/th>\n[latex]\\frac{d}{dx}f(x)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
[latex]{\\text{sinh}}^{-1}x[\/latex]<\/td>\n[latex]\\frac{1}{\\sqrt{1+{x}^{2}}}[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{cosh}}^{-1}x[\/latex]<\/td>\n[latex]\\frac{1}{\\sqrt{{x}^{2}-1}}[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{tanh}}^{-1}x[\/latex]<\/td>\n[latex]\\frac{1}{1-{x}^{2}}[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{coth}}^{-1}x[\/latex]<\/td>\n[latex]\\frac{1}{1-{x}^{2}}[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{sech}}^{-1}x[\/latex]<\/td>\n[latex]\\frac{-1}{x\\sqrt{1-{x}^{2}}}[\/latex]<\/td>\n<\/tr>\n
[latex]{\\text{csch}}^{-1}x[\/latex]<\/td>\n[latex]\\frac{-1}{|x|\\sqrt{1+{x}^{2}}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n
\n

Note that the derivatives of [latex]{\\text{tanh}}^{-1}x[\/latex] and [latex]{\\text{coth}}^{-1}x[\/latex] are the same. Thus, when we integrate [latex]1\\text{\/}(1-{x}^{2}),[\/latex] we need to select the proper antiderivative based on the domain of the functions and the values of [latex]x.[\/latex]<\/p>\n<\/section>\n

\n

 integral formulas for inverse hyperbolic functions<\/h3>\n

[latex]\\begin{array}{cccccccc}\\hfill \\displaystyle\\int \\frac{1}{\\sqrt{1+{u}^{2}}}du& =\\hfill & {\\text{sinh}}^{-1}u+C\\hfill & & & \\hfill \\displaystyle\\int \\frac{1}{u\\sqrt{1-{u}^{2}}}du& =\\hfill & \\text{\u2212}{\\text{sech}}^{-1}|u|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\frac{1}{\\sqrt{{u}^{2}-1}}du& =\\hfill & {\\text{cosh}}^{-1}u+C\\hfill & & & \\hfill \\displaystyle\\int \\frac{1}{u\\sqrt{1+{u}^{2}}}du& =\\hfill & \\text{\u2212}{\\text{csch}}^{-1}|u|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\frac{1}{1-{u}^{2}}du& =\\hfill & \\bigg\\{\\begin{array}{c}{\\text{tanh}}^{-1}u+C\\text{ if }|u|<1\\hfill \\\\ {\\text{coth}}^{-1}u+C\\text{ if }|u|>1\\hfill \\end{array}\\hfill & & & & & \\end{array}[\/latex]<\/p>\n<\/section>\n

\n

Evaluate the following derivatives:<\/p>\n

    \n\t
  1. [latex]\\frac{d}{dx}({\\text{sinh}}^{-1}(\\frac{x}{3}))[\/latex]<\/li>\n\t
  2. [latex]\\frac{d}{dx}{({\\text{tanh}}^{-1}x)}^{2}[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167793949787\"]Show Solution[\/reveal-answer]
    \n[hidden-answer a=\"fs-id1167793949787\"]\n\n

    Using the formulas in the table on derivatives of the inverse hyperbolic functions and the chain rule, we obtain the following results:<\/p>\n

      \n\t
    1. [latex]\\frac{d}{dx}({\\text{sinh}}^{-1}(\\frac{x}{3}))=\\frac{1}{3\\sqrt{1+\\frac{{x}^{2}}{9}}}=\\frac{1}{\\sqrt{9+{x}^{2}}}[\/latex]<\/li>\n\t
    2. [latex]\\frac{d}{dx}{({\\text{tanh}}^{-1}x)}^{2}=\\frac{2({\\text{tanh}}^{-1}x)}{1-{x}^{2}}[\/latex]<\/li>\n<\/ol>\n

      [\/hidden-answer]<\/p>\n<\/section>\n

      \n

      Evaluate the following integrals:<\/p>\n

        \n\t
      1. [latex]\\displaystyle\\int \\frac{1}{\\sqrt{{x}^{2}-4}}dx,\\text{}x>2[\/latex]<\/li>\n\t
      2. [latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-{e}^{2x}}}dx[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167793371557\"]Show Solution[\/reveal-answer]
        \n[hidden-answer a=\"fs-id1167793371557\"]\n\n
          \n\t
        1. [latex]\\displaystyle\\int \\frac{1}{\\sqrt{{x}^{2}-4}}dx={\\text{cosh}}^{-1}(\\frac{x}{2})+C[\/latex]<\/li>\n\t
        2. [latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-{e}^{2x}}}dx=\\text{\u2212}{\\text{sech}}^{-1}({e}^{x})+C[\/latex]<\/li>\n<\/ol>\n

          Watch the following video to see the worked solution to this example.<\/p>\n