Question

if y=(x^2-1)^n prove that (x^2-1)yn+2 + 2xyn+1 - n(n+1)yn =0 differentiating (n+1) times using Leibnitz theorem
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Answer 1

The answer for first order derivative will be :

n∗(x^2–1)(n−1)∗2x

while doing second time don’t consider “n” as it is free from x. The answer will be :

n∗(x^2–1)(n−1)∗2+2x∗(n−1)∗(x2−1)(n−2)∗2x