# Given the plot of $(v, f(v))$ how to scale horizontal axis with $v \to 87\sqrt{-2\log(1 - v)})$, while keeping the vertical axis unchanged?

In my previous post @herbertfederer provided a script to plot $$\frac{\partial C}{\partial u}\Big|_{u=0.5}$$ where $$C(u,v)= \min\left[ u^{0.0325}v, uv^{-0.1018} \right]$$,

c[u_, v_] := Min[u^0.0325 v, u v^-0.1018]
Plot[D[c[u, v], u] /. u -> .5 // Evaluate, {v, 0, 1}]


Here I modified the range of $$v$$ to $$(0,1)$$. Now it plots a function say $$f(v)$$ for the horizontal $$v$$ values. I need to assign $$87\sqrt{-2\log(1 - v)}$$ to $$v$$ and plot the function based on this new horizontal axis. So I tried,

c[u_, v_] := Min[u^0.0325 v, u v^-0.1018]
Plot[D[c[u, v], u] /. u -> 0.5 /. v -> (87*Sqrt[-2*Log[1 - v]]) // Evaluate, {v, 0, 0.9999}]


I choose "{v, 0, 0.9999}" in order to $$\log(1-v)$$ be defined. But after running the script I got errors. Can you help me to fix the issue?

• Try ParametricPlot Commented Dec 16, 2023 at 8:47
• @UlrichNeumann Can you elaborate on that? I'm new to mathematica. Commented Dec 16, 2023 at 9:09
• Did you modify the question? I remember v->h[x]? Commented Dec 16, 2023 at 12:10
• @UlrichNeumann Yes I realized my explanation was not accurate and found a better way to explain the problem. Commented Dec 16, 2023 at 16:56

$Version (* "14.0.0 for Mac OS X x86 (64-bit) (December 13, 2023)" *) c[u_, v_] := Min[u^0.0325 v, u v^-0.1018]  Using ParametricPlot as suggested by Ulrich Neumann ParametricPlot[{(87*Sqrt[-2*Log[1 - v]]), Derivative[1, 0][c][0.5, v]}, {v, 0, 1}, AspectRatio -> 1/GoldenRatio, ColorFunction -> (ColorData["Rainbow"][#3] &), PlotLegends -> BarLegend[{"Rainbow", {0, 1}}, LegendLabel -> Style[v, 14], LegendMarkerSize -> 220], Frame -> True, FrameLabel -> (Style[#, 14] & /@ {(87*Sqrt[-2*Log[1 - v]]), HoldForm[Derivative[1, 0][c][0.5, v]]})]  The new v looks like: Plot[-87*Sqrt[-2*Log[1 - v]], {v, 0, 0.9999}]  Now rise this to the power of : -0.1018: Table[(-87*Sqrt[-2*Log[1 - v]])^-0.1018, {v, 0, 0.9, 0.1}] {ComplexInfinity, 0.6522 - 0.215998 I, 0.627758 - 0.207903 I, 0.612949 - 0.202998 I, 0.601844 - 0.199321 I, 0.592567 - 0.196248 I, 0.584208 - 0.19348 I, 0.576145 - 0.19081 I, 0.567696 - 0.188011 I, 0.55744 - 0.184615 I}  You see that the new v is complex. • Sorry, I realized I wrongly put an extra minus sign in the expression. It should be 87*Sqrt[-2*Log[1 - v]]. Thanks for detecting it. Commented Dec 16, 2023 at 17:04 • For example assume the map of$v=0.4$with above function is 90 and assume the original graph of partial derivative gives$f(0.4)= 0.3$I'm looking for a plot that gives$f(90)=0.3$. And this pattern holds for other value of$v\in[0,1)$. And I want to scale only horizontal axis by the pre-defined function. Commented Dec 16, 2023 at 17:32 • I am a bit confused because you have 2 variables: u, v. Are you looking for: f'( Inverse(f)(x)) ? Commented Dec 16, 2023 at 19:11 • The variable u in the original function C(u,v) is only used for partial derivative and then plugging a specific u, i.e$\large\frac{\partial C}{\partial u}\Big|_{u=0.5}\$ And this gives a function based on only the variable v. Now I want to transform the horizontal axis of the plot of this function the way I stated in my previous comment. Commented Dec 16, 2023 at 21:19
From your comment, I take it that you are looking for:  f'( Inverse(f)(x) ).
Plot[c[0.1, x], {x, 0, 1}, PlotRange -> All]