Tangent Slope at Point y=\cosh x = 1

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SUMMARY

The discussion focuses on finding the point on the curve defined by the equation y = cosh(x) where the tangent slope equals 1. The solution involves using the derivative of the hyperbolic sine function, leading to the equation 1 = sinh(x) * dy/dx. The final answer is determined to be x = ln(1 + √2), providing a clear resolution to the problem posed.

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  • Understanding of hyperbolic functions, specifically cosh and sinh.
  • Knowledge of derivatives and their application in finding slopes of curves.
  • Familiarity with logarithmic functions and their properties.
  • Basic calculus concepts, including the chain rule and implicit differentiation.
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  • Study the properties and applications of hyperbolic functions in calculus.
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bard
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at what point on the curve [tex]y=\cosh x[/tex] does the tangent have slope 1


I have no idea how to approach this problem

my work

[tex]1=\sinh x\frac{dy}{dx}[/tex]

[tex]\frac{1}{sinh x}=\frac{dy}{dx}[/tex]
 
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It would probably help to you use the definitions.

For example:
[tex]\sinh{x}= \frac{e^{x}-e^{-x}}{2}[/tex]
 
yah I found the answer as x=ln[1+sqt2]----thanks for your help
 

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