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

$\text{csch} \theta \text{coth} \theta (\ln csch \theta)$
Since, $\dfrac{d}{dx} (csch x)=\text{-csch} x \text{coth} x$ As we are given that $y=csch \theta (1-\ln csch \theta)$ Then, on differentiating , we have: $\dfrac{dy}{d \theta}=csch \theta[\dfrac{-1}{csch \theta}( -csch \theta coth \theta)]+(1-\ln csch \theta)=(csch \theta coth \theta)[1-(1-\ln csch \theta)]=\text{csch} \theta \text{coth} \theta (\ln csch \theta)$