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

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

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Answer 1 · 2015-04-02T15:31:15

let $y = cosh (ln x)$ $y = cosh(lnx)$
$\Rightarrow y = \frac{1}{2} \cdot (e^{ln x} - e^{- ln x})$ $=> y = 1/2*(e^(lnx) - e^(-lnx))$

$= \frac{1}{2} \cdot (e^{ln x} + e^{{ln x}^{- 1}})$ $= 1/2*(e^(lnx) + e^(lnx^-1))$

$= \frac{1}{2} (x + x^{- 1})$ $= 1/2(x + x^-1)$

$\frac{d y}{d x} = \frac{1}{2} (1 + (- 1) \cdot x^{- 2}) = \frac{1}{2} (\frac{x^{2} - 1}{x^{2}}) = \frac{x^{2} - 1}{2 x^{2}}$ $(dy)/(dx) = 1/2(1 + (-1)*x^-2) = 1/2((x^2 - 1)/x^2) = (x^2 - 1)/(2x^2)$

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

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