# Finding the derivative of InverseFunction[ ] ?

Posted 22 days ago
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Posted 22 days ago
 You can start with the basic relation f[(f^-1)[x]] == x and take its derivative: f = -((Sqrt[#1] (-E^(2 C[1]) + #1) + E^(3 C[1]) ArcSin[E^-C[1] Sqrt[#1]] Sqrt[1 - E^(-2 C[1]) #1])/Sqrt[E^(2 C[1]) - #1]) &; basicRelation = (f[fInverse[t + C[2]]] == t + C[2]); Solve[D[basicRelation, t], fInverse'[t + C[2]]] 
Posted 22 days ago
 Thank you! By the way, I have another problem (not related to the above one) to ask you: I used the NDSolve to solve a set of ordinary differential equations in a very narrow interval of the independent variable(say t) and it failed. Actually I managed to compute the values of functions in the equations at each t with very high working precision, but the values of functions weren't calculated with such high working precision in the NDSolve solving. It seemed I was solving at a single point. When I made the interval large, it worked. How can I solve the problem? Need I write my own differential equation solving program instead of NDSolve?
Posted 22 days ago
 Perhaps you can rescale the parameter t in the equation so that it ranges over a larger interval.
 You can always rescale the independent variable, for example this way: independentVariableChange = (a*t + b == x) dependentVariableChange = (g[t] == f[x]) dependentVariableChange /. Solve[independentVariableChange, x][[1]] % /. Solve[independentVariableChange, t][[1]] changeOfVariables = f -> (Function[x, g[t]] /. Solve[independentVariableChange, t][[1]]) originalDiffEq = f''[x] - f'[x] == x*f[x] originalDiffEq /. changeOfVariables % /. Solve[independentVariableChange, x][[1]]