I was trying to find the $n$th derivative of the function $$y=e^x\tanh^{-1}x$$
I managed to find a formula for the $n$th derivative by inspection that relates the $n$th derivatives with its lower order derivatives : $$(x^2-1)(y^{(n)}-y^{(n-1)})+2x(n-1)(y^{(n-1)}-y^{(n-2)})+n(n-1)(y^{(n-2)}-y^{(n-3)})+e^x=0$$
This applies only for values of $n \geq 3$
But is there a way to find the $n$th derivative explicitly in terms of $x$ and $n$ only?
Hint.
The general Leibniz rule says that $$ (fg)^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}g^{(k)} $$
Now let $f(x)=e^x$, $g(x)=\tanh^{-1}(x)$. So all you need is to find the $n$-th derivative for $g$.
Note that $$ g'(x)=\frac{1}{1-x^2}=\frac12\left(\frac{1}{x+1}-\frac{1}{x-1}\right) $$ By linearity, you only need to find the $(n-1)$-th derivative for the following two functions: $$ \frac{1}{x+1},\quad \frac{1}{x-1}, $$ which should be straightforward by induction.