Practice your identities. Don't stop because you have an answer. Play with it. Convert it into something else. See if your answer actually matches the answer in the back of the book. Try to recognize things like double angle formulas.
Practice your identities. Don't stop because you have an answer. Play with it. Convert it into something else. See if your answer actually matches the answer in the back of the book. Try to recognize things like double angle formulas.
I played with my original answer substituting the correct trigonometric hyperbolic identities and ended up with y' = 2/[sinh^2(x/2)]. However, I don't know how to simplify further to reach the textbook answer of csch(x). Can someone take it from here to find the textbook answer csch(x)?
$\displaystyle \begin{align*} y' & = \dfrac{ \text{sech}^2\left( \frac{x}{2} \right) }{ 2 \tanh\left( \frac{x}{2} \right) } \\ & = \dfrac{ \left( \dfrac{ 1 }{ \cosh^2\left( \frac{x}{2} \right) } \right) }{ 2 \dfrac{ \sinh\left( \frac{x}{2} \right) }{ \cosh\left( \frac{x}{2} \right) } } \cdot \dfrac{\cosh^2\left( \frac{x}{2} \right) }{ \cosh^2\left( \frac{x}{2} \right) } \\ & = \dfrac{ 1 }{ 2\sinh\left( \frac{x}{2} \right)\cosh\left( \frac{x}{2} \right) } \\ & = \dfrac{1}{\sinh(x)} \\ & = \text{csch}(x) \end{align*}$