Answer
$y'=-\sec h x\tan h x(2+\ln \sec h x)$
Work Step by Step
$\DeclareMathOperator{\sech}{sech}$
$y'=\frac{d}{dx}\sech x (1+\ln\sech x)$
Using the product rule:
$y'=-\sech x\tanh x(1+\ln\sech x)+\sech x(\frac{d}{dx}(1+\ln\sech x))$
$y'=-\sech x\tanh x(1+\ln\sech x)+\sech x(\frac{d}{dx}(\ln\sech x))$
Using the chain rule:
$y'=-\sech x\tanh x(1+\ln\sech x)+(\sech x)(\frac{d\ln\sech x}{d\sech x}\times\frac{d\sech x}{dx})$
$=-\sech x\tanh x(1+\ln\sech x)+(\sech x)(\frac{1}{\sech x}\times(-\sech x\tanh x))$
$=-\sech x\tanh x(1+\ln\sech x)+(\sech x)(-\tanh x)$
$=-\sech x\tanh x(1+\ln\sech x)-\sech x\tanh x(1)$
$=-\sech x\tanh x(2+\ln\sech x)$
