# How to Make a change of variables

I have the following expresion

$$\hat{M}^2=-\left[\frac{1}{\sin\beta}\frac{\partial}{\partial\beta}\sin\beta\frac{\partial}{\partial\beta}+\frac{1}{sin^2\beta}\left(\frac{\partial^2}{\partial\alpha^2}+\frac{\partial^2}{\partial\gamma^2}\right)-2\frac{cot\beta}{sin\beta}\frac{\partial^2}{\partial\alpha\partial\gamma}\right]$$

In my problem i don't want to work with the angle $\beta$, instead i just want to work with the $\cos\beta$ (basically due to the convergence of Legendre functions with respect to spherical armonics), so i'm trying to make a transformation rule for the partial derivative. For instance, $$q=\cos\beta \quad; usin=\sin\beta=\sqrt{1-q^2} \quad and\quad dq=-\sin\beta d\beta$$

My expression for $\hat{M^2}$ is now: $$\hat{M}^2=-\left[\frac{\partial}{\partial q}usin^2\frac{\partial}{\partial q }+ \frac{1}{usin^2}\left(\frac{\partial^2}{\partial\alpha^2}+\frac{\partial^2}{\partial\gamma^2}\right)-2\frac{q}{usin^2}\frac{\partial^2}{\partial\alpha\partial\gamma}\right]$$

Subscript[w, 1] = f[Subscript[r, 1], Subscript[r, 2], α, β, γ, Subscript[Θ, 12]]

M = -(1/Sin[β] D[(Sin[β] D[Subscript[w, 1], β]), β] + 1/Sin[β]^2 (D[Subscript[w, 1],
{α, 2}] + D[Subscript[w, 1], {γ, 2}]) - 2 Cot[β]/Sin[β] D[Subscript[w, 1], α, γ])

I have written a few lines of code, however i don't know how to exactly proceed with the transformation.

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$Assumptions = usin > 0 so expressions like Sqrt[usin^2] will be simplified to usin. M2 /. q^2 -> 1 - usin^2 // Simplify The output is your desired result but in expanded form. I'm not sure how to make it in the exact same form as yours. - Many thanks for your help @Kuba it was really usefull, both the chain of variables and the formated output for derivatives. I'm wondering if i could apply the same procedure with more than one angle, for instance also for γ, Can i do this: M2 /. f -> (f[#, #2, #3, Cos[#4], Cos[#5], #6] &) /. β -> ArcCos[q] /. γ -> ArcCos[s];$Assumptions = usinq > 0 && usins > 0 M2 /. q^2 -> 1 - usinq^2 /. s^2 -> 1 - usins^2 // ExpandAll. – shadraws Mar 29 '14 at 0:04