Problem source

Let $y = \ln (x^2-1)\,$, where $x  >1$,  and let $r$ and $\theta$ be functions of $x$ determined by $r= \sqrt{x^2-1}$ and $\coth\theta=  x$. Show that 
\[
\frac {\d y}{\d x} = \frac {2\cosh \theta}{r}
\text{ and }
\frac {\d^2 y}{\d x^2} = -\frac {2 \cosh 2\theta}{r^2}\,,
\]
and find an expression in terms of $r$ and $\theta$ for $\dfrac {\d^3 y}{\d x^3}\,$.
 
Find, with proof, a similar formula for $\dfrac{\d^n y}{\d x^n}$ in terms of $r$ and $\theta$.

Solution source

\begin{align*}
&& y &= \ln(x^2 -1) \\
&& r &= \sqrt{x^2-1} \\
&& \coth \theta &= x \\
&& r &= \sqrt{\coth^2 \theta - 1} = \sqrt{\textrm{cosech}^2 \theta} = \textrm{cosech} \theta 
\\
&& \frac{\d y}{\d x} &= \frac{2x}{x^2-1} \\
&&&= \frac{2 \coth \theta}{r^2} \\
&&&= \frac{2 \cosh \theta}{\sinh \theta \cdot r \cdot \textrm{cosech} \theta } \\
&&&= \frac{2 \cosh \theta}{r } \\
\\
&& \frac{\d^2 y}{\d x^2} &= \frac{2(x^2-1)-4x^2}{(x^2-1)^2} \\
&&&= \frac{-2(1+x^2)}{r^2 \textrm{cosech}^2 r} \\
&&&= -\frac{2(1 + \coth^2 \theta) \sinh^2 \theta}{r^2} \\
&&&= -\frac{2(\sinh^2 \theta + \cosh^2 \theta)}{r^2} \\
&&&= -\frac{2 \cosh 2 \theta}{r^2} \\
\\
&& \frac{\d^3 y}{\d x^3} &= \frac{-4x(x^2-1)^2-(-2x^2-2)\cdot2(x^2-1)\cdot 2x}{(x^2-1)^4} \\
&&&= \frac{-4x(x^2-1)+8x(x^2+1)}{(x^2-1)^3}\\
&&&= \frac{4x^3+12x}{(x^2-1)^3} \\
&&&=\frac{\sinh^3 \theta (4\coth^3 \theta + 12\coth \theta )}{r^3} \\
&&&=\frac{4\cosh^3 \theta + 12\cosh \theta \sinh^2 \theta}{r^3} \\
&&&= \frac{4 \cosh 3 \theta}{r^3} \\
\end{align*}

Claim: $\frac{\d^n y}{\d x^n} = (-1)^{n+1}\frac{2(n-1)!\cosh n \theta}{r^n}$
Proof: By induction. Base cases already proven

\begin{align*}
\frac{\d r}{\d x} &= \frac{x}{\sqrt{x^2-1}} = \frac{\coth \theta}{\textrm{cosech} \theta} = \cosh \theta
\\
\frac{\d \theta}{\d x} &= - \sinh^2 \theta \\
\\
\frac{\d^{n+1} y}{\d x^{n+1}} &= (-1)^{n+1}(n-1)!\frac{\d}{\d x} \left ( \frac{2\cosh n \theta}{r^n}\right) \\
&= (-1)^{n+1}\frac{2 n \sinh n \theta \cdot r^n \cdot \frac{\d \theta}{\d x}- 2\cosh n \theta \cdot nr^{n-1} \frac{\d r}{\d x} }{r^{2n}} \\
&= (-1)^{n+2}\frac{2n( \cosh n \theta\cosh \theta + r\sinh n \theta \sinh^2 \theta) }{r^{n+1}} \\
&= (-1)^{n+2}n!\frac{2\cosh(n+1) \theta }{r^{n+1}} \\
\end{align*}


We can think of this as $\ln(x^2-1) = \ln(x+1)+\ln(x-1)$ and also note $x \pm 1 = \coth \theta \pm  1 = \frac{\cosh \theta \pm \sinh \theta}{\sinh \theta} = \frac{e^{\pm \theta}}{\sinh \theta}$

\begin{align*}
&& \frac{\d^n}{\d x^n} \ln(x^2-1) &= (n-1)!(-1)^{n-1} \left ( \frac{1}{(x+1)^n} + \frac{1}{(x-1)^n} \right) \\
&&&= (-1)^{n-1}(n-1)! \left ( \frac{\sinh^n \theta}{e^{n\theta}} + \frac{\sinh^n \theta}{e^{-n\theta}} \right) \\
&&&= (-1)^{n-1} (n-1)!2\cosh n \theta \cdot \sinh^n \theta \\
&&&= (-1)^{n-1}(n-1)! \frac{2 \cosh n \theta }{r^n}
\end{align*}