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March 28, 2015

March 28, 2015

Posted by **Heather** on Thursday, October 4, 2012 at 1:10pm.

- Calculus 2 -
**Reiny**, Thursday, October 4, 2012 at 1:53pmlet y = ln [ x + (x^2 - 1)^(1/2) ]

dy/dx = 1/(x + (x^2 - 1)^(1/2) * (1 + (1/2)(x^2 - 1)^(-1/2) (2x) )

= 1/(x + √(x^2 - 10 ) * (1 + x/√(x^2 - 1) )

= 1/(x + √(x^2 - 10 ) * (√(x^2 -1) + x)/√(x^2-1)

let's multiply top and bottom by x - √(x^2 - 1) , thus rationalizing the denominator in the first part

= 1/(x + √(x^2 - 10 ) * (√(x^2 -1) + x)/√(x^2-1) * [x - √(x^2 - 1)]/[x - √(x^2 - 1)]

= (x^2 - x^2 + 1)/( (√x^2 - 1)(x^2 - x^2 + 1) )

= 1/√(x^2 - 1)

Whewww!

- Calculus 2 -
**Steve**, Thursday, October 4, 2012 at 2:10pmyou can make things a little less complicated if you recognize that

ln(x+√(x^2-1)) = arccosh(x)

d/dx arccosh(x) = 1/√(x^2-1)

Reiny's excellent work shows how it's done.

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