Use S=(1+(M/m))^-1 to write an expression for delta S in terms of delta m and delta M.?

S = [1 + (M/m)]⁻¹
S = [(m/m) + (M/m)]⁻¹
S = [(m + M) / m]⁻¹
S = m / (M + m) ...... [1]

When m is changed to (m + Δm) and M is changed to (M + ΔM), S is changed to (S + ΔS).
S + ΔS = (m + Δm) / [(M + ΔM) + (m + Δm)]
S + ΔS = (m + Δm) / (M + m + ΔM + Δm) ...... [2]

[2] - [1] :
ΔS = [(m + Δm)/(M + m + ΔM + Δm)] - [m/(M + m)]
ΔS = {(m+ Δm)(M + m) / [ (M + m)(M + m + ΔM + Δm)]} - {m(M + m + ΔM + Δm) / [(M + m)(M + m + ΔM + Δm)]}
ΔS = {(mM+ m² + MΔm + mΔm) / [ (M + m)(M + m + ΔM + Δm)]} - {(mM + m² + mΔM + (Δm)²) / [(M + m)(M + m + ΔM + Δm)]}
ΔS = [mM+ m² + MΔm + mΔm - mM - m² - mΔM - (Δm)²] / [(M + m)(M + m + ΔM + Δm)]
ΔS = [MΔm + mΔm - mΔM - (Δm)²] / [(M + m)(M + m + ΔM + Δm)]

Rewrite the eqn
S= 1/[1+(M/m)]
If we increase M, (keep m as say =1) we see that S = 1/; 1/3; 1/4, etc ,so as M increases, S decreases and vice versa
Do the same with m and we get 1/2; 1/1.5; 1/1.33, etc and we see that S increases as m increases and vice versa.