Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 24

$4$

Hyperbolic functions identities are: $ \sinh x= \dfrac{e^x-e^{-x}}{2}$ Thus, $\text{csch} x=\dfrac{1}{\sinh x}=\dfrac{2}{e^x -e^{-x}}$ Given: $y=(4x^2 -1) \text{csch}(\ln 2x)$ Thus, $y=(4x^2 -1)\dfrac{2}{e^{\ln 2x} - e^{-\ln 2x}}= (4x^2 -1)\dfrac{2}{[e^{\ln 2x}+e^{\ln (2x)^{-1}}]}=(4x^2 -1)\dfrac{4x}{4x^2-1}=4x$ Hence, $\dfrac{dy}{dx}=4$