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\begin{document}

\begin{questions}

\question 

\begin{parts}


\part Most identities about regular trig functions are related to the fundamental identity $\sin^2(x) + \cos^2(x) = 1$. What is the fundamental identity for hyperbolic trig functions? Show that this identity holds by using the definitions 
$$
\sinh(x) = \frac{e^x - e^{-x}}{2} \quad \hbox{ and } \quad \cosh(x) = \frac{e^x + e^{-x}}{2}.
$$

% cosh^2 - sinh^2 = 1

\part Divide the identity from (a) by $\sinh^2$ or $\cosh^2$ to get two new identities.

% coth^2 - 1 = csch^2
% 1 - tanh^2 = sech^2

\part Use Implicit Differentiation to show that the derivative of $\tanh^{-1}(x)$ is $\frac{1}{1-x^2}$. You will need to make use of the identity in $(a)$ or one of the identities in (b). [Hint: If you're not sure how to do this problem, you may use part (d) as a roadmap.]

\part Below is an attempt to compute the derivative of the inverse hyperbolic cosecant function $\operatorname{csch}^{-1}(x)$. There is a subtle difference between the formula obtained below and the one given to you in class / in the textbook. Locate the error in this argument and fix it.

% tanh(y) = x
% sech^2(y) y' = 1
% y = 1 / sech^2(y)
% Since tanh=x and 1-tanh^2 = sech^2, sech^2 = 1-x^2
% Therefore y = 1/ 1-x^2

\vspace{6pt}
Let $y=\operatorname{csch}^{-1}(x)$. Then $\operatorname{csch}(y)=x$, i.e. $1= x \sinh(y) $. Hence
$$
0 = x \cosh(y) \frac{dy}{dx} + \sinh(y).
$$
Therefore
$ \displaystyle
\frac{dy}{dx} = -\frac{\sinh(y)}{x \cosh(y)}.
$
Since $\sinh(y) = x^{-1}$,
$$
\cosh(y) = \sqrt{1+x^{-2}} = x^{-1} \sqrt{x^2+1}.
$$
So finally we get
$$
\frac{dy}{dx} = - \frac{x^{-1}}{x \cdot x^{-1} \sqrt{x^2+1}} = - \frac{1}{x \sqrt{x^2+1}}.
$$

\end{parts}


\end{questions}

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