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

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Apr 2, 2015

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)#

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