How do you find the derivative of $cosh(ln x)$ ?

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Answer 1 · 2015-04-02T15:31:15

let $y = cosh(lnx)$
$=> y = 1/2*(e^(lnx) - e^(-lnx))$

$= 1/2*(e^(lnx) + e^(lnx^-1))$

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

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

How do you find the derivative of cosh(ln x)cosh(ln x)?