# How do you find derivative of f(x)=cosh(lnx)?

Aug 9, 2015

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{1}{2} \left(1 - \frac{1}{x} ^ 2\right)$

#### Explanation:

$f \left(y\right) = \cosh y = \frac{{e}^{y} + {e}^{-} y}{2}$

Taking $y = \ln x$
$\cosh \left(\ln x\right) = \frac{x + \frac{1}{x}}{2}$
Taking the derivative with respect to x:
$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{1}{2} \left(1 - \frac{1}{x} ^ 2\right)$