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Prove Equation 5 using (a) the method of Example 3 and (b) Exercise 18 with $ x $ replaced by $ y. $
a. $y=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$b. $y=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 11
Hyperbolic Functions
Derivatives
Differentiation
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Okay. We know that X is either y minus C to the native. Lie over each of the y plus e to the negative. Why? Therefore, we have X minus one each of the two. Why is equivalent to negative one plus tax? Therefore, either too. Why one plus ax over one minus x the natural log of each of the two. Why the natural of one plus acts over one minus X? Because we're taking the nuptial both sides drink. This gives us why is 1/2 natural look of one plus acts for one minus X Therefore could just consider this to be inverse tangent Kovacs and we know that X is between negative one positive one.
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