Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 29

$$y'= \frac{e^{\cosh^{-1}x} }{\sqrt{x^2-1}}.$$

Work Step by Step

Recall that $(\cosh^{-1} x)'=\dfrac{1}{\sqrt{x^2-1}}$ Recall that $(e^x)'=e^x$ Since $y=e^{\cosh^{-1}x}$, then the derivative, by using the chain rule, is given by $$y'= e^{\cosh^{-1}x}({\cosh^{-1}x})'=\frac{e^{\cosh^{-1}x} }{\sqrt{x^2-1}}.$$

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