## University Calculus: Early Transcendentals (3rd Edition)

$2$
Use the hyperbolic functions formulae such as: $\cosh x= \dfrac{e^x+e^{-x}}{2}$ Thus, $\text{sech} x=\dfrac{1}{\cosh x}=\dfrac{2}{e^x+e^{-x}}$ We are given $y=(x^2+1) \text{sech}(\ln x)$ or, $y=(x^2+1)\dfrac{2}{e^{\ln x}+e^{-\ln x}}=(x^2+1)\dfrac{2}{e^{\ln x}+e^{\ln x^{-1}}}=(x^2+1)\dfrac{2x}{x^2+1}=2x$ Now, $\dfrac{dy}{dx}=2$