Tuesday

November 25, 2014

November 25, 2014

Posted by **mary** on Tuesday, May 3, 2011 at 3:35pm.

- calculus -
**Reiny**, Tuesday, May 3, 2011 at 4:48pmf(x) = ln(e^x - e^-x)

f'(x) = (e^x + e^-x)/(e^x - e^-x)

f''(x) = [(e^x - e^-x)(e^x - e^-x) - (e^x + e^-x)(e^x + e^-x)]/(e^x - e^-x)^2

= [(e^x - e^-x)^2 - (e^x + e^-x)^2]/(e^x - e^-x)^2

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