`y = (e^u - e^-u)/(e^u + e^-u)` Find the derivative of the function.

Textbook Question

Chapter 3, 3.4 - Problem 28 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsenviro | College Teacher | (Level 1) Educator Emeritus

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Using the quotient rule of derivatives:

`y' = [d/(du) (e^u-e^(-u)) * (e^u+e^(-u)) - (e^u-e^(-u)) d/(du) (e^u +e^(-u))]/(e^u+e^(-u))^2`

` ` `= [(e^u -e^(-u) *(-1))*(e^u+e^(-u)) - (e^u-e^(-u))*(e^u+e^(-u) *(-1))]/(e^u+e^(-u))^2`

`=[(e^u + e^(-u))*(e^u+e^(-u))- (e^u-e^(-u))*(e^u-e^(-u))]/(e^u+e^(-u))^2`

`= [(e^u+e^(-u))^2-(e^u-e^(-u))^2] /(e^u+e^(-u))^2`

`= [(e^u +e^(-u)+e^u-e^(-u)) *(e^u+e^(-u)-e^u+e^(-u))]/(e^u+e^(-u))^2`

`= (4e^ue^(-u))/(e^u+e^(-u))^2 `

`= 4/(e^u+e^(-u))^2`

Hope this helps.

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