I have the following function: $$ y(x)=\frac{Ax}{Bx^{1/4}\sinh{(\frac{Bx^{1/4}}{2})}}\left(\cosh{(\frac{Bx^{1/4}}{2})}-1 \right), $$ where $A=112.941$ and $B=18.6588$. I want to find the differential of the inverse function: $$ x'(y)=\frac{dx}{dy}. $$ Since the function is transcendental, I cannot simply invert the function and find an explicit expression for $x(y)$. So I used implicit differentiation in Mathematica, like so:
D[y - ((A*x[y])/(B*(x[y])^(1/4)*Sinh[B*(x[y])^(1/4)/2]))*(Cosh[
B*(x[y])^(1/4)/2] - 1), y],
and solved for $x'(y)$. FullSimplify gives
(16 B Cosh[1/4 B x[y]^(1/4)]^2 x[y]^(1/4))/(
6 A Sinh[1/2 B x[y]^(1/4)] + A B x[y]^(1/4)),
which is a plottable function.
It has been a while since I last dealt with inverse functions and there is no easy way to heuristically varify the result. Is this approach correct at all? A confirmation from fellow Mathematica users would be greatly appreciated.
P.S.: I am sorry if this is the wrong stackexchange for this question, but others seemed even less fitting. Thank you for your time and have a great weekend!