Really annoying derivatives

[K.I.L.E.R]

L/Ls = (M/Ms)^(7/2)

Nice equation to turn into a Taylor series, right?
WRONG!

f(x) = (M/Ms)^(7/2)
f'(x) = ((7/2)*((M/Ms)^(5/2)))*((Ms-M)/(Ms*Ms))

Prove that derivative was derived successfully:
f(x)/f'(x) = 1

f(x) = (M/Ms)
Function Added
f = ((7/2)*((M/Ms)^(5/2)))*((Ms-M)/(Ms*Ms))
Function Added

f(1)/f(1)
1

This one added to see if x, y values do anything to function(I don't trust Microsoft)
f(3)/f(2)
1

So I got a derivative of the original function, however to get an appropriate Taylor series I need at least 3 derivatives.

So I now need to take the derivative of f'(x), as you can see I know need multiple chain rules, it's hell.
Is there ANY mathematical method that will speed this process of obtaining derivatives?

 

[Fred]

It would help if you labeled your functions. I take it M is a function of X, but what is Ms? What is L and Ls.. And where do you get functions of y???

 

[K.I.L.E.R]

x and y are dummy variables, intended to play nice with Powertoys calculator.

Ms = Solar Mass
Ls = Solar luminosity

M = mass of another star
L = luminosity of another star

 

[KimB]

Well, you should be holding solar mass and luminosity fixed, so your derivatives should all be with respect to M. In those equations, you appear to be incorrectly attempting to take derivatives with respect to both M and Ms.

Anyway, this sort of equation is going to be very difficult to take a Taylor expansion of if you attempt to expand the function about zero. You might do better if you instead expand about, say, M=Ms.

 

[K.I.L.E.R]

I didn't see that, thanks.
I'm reading the sites you and Neeyik gave me and I'm taking a Taylor series of every formula there is.