Answer
$4$
Work Step by Step
Use the hyperbolic functions formulae such as: $ \sinh x= \dfrac{e^x-e^{-x}}{2}$ Thus, $\text{csch} x=\dfrac{1}{\sinh x}=\dfrac{2}{e^x -e^{-x}}$
We are given $y=(4x^2 -1) \text{csch}(\ln 2x)$
or, $y=(4x^2 -1)\dfrac{2}{e^{\ln 2x} - e^{-\ln 2x}}= (4x^2 -1)\dfrac{2}{e^{\ln 2x}+e^{\ln (2x)^{-1}}}=(4x^2 -1)\dfrac{4x}{4x^2-1}=4x$
Now, $\dfrac{dy}{dx}=4$
