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प्रश्न
Find nth derivative of `1/(x^2+a^2.`
योग
उत्तर
y = `1/(x^2+a^2`
`y = 1/((x+ai)(x-ai))`
Let `1/((x+ai)(x-ai))="A"/(x+ai)+B/(x-ai)`
`1 = "A"(x-ai)+"B"(x+ai)`
Put x = ai
1 = B(2ai)
B = `1/(2ai)`
Put x = -ai, we get
`"A"=-1/(2ai)`
`therefore y=(1/(-2ai))/(x+ai)+(1/(2ai))/(x-ai)`
`therefore y = 1/(2ai)[1/(x+ai)-1/(x-ai)]`
Diff n times, we get
`y_n=1/(2ai)[((-1)^n n!)/((x-ai)^(n+1))-((-1)^n n!)/(x+ai)^(n+1)].`
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nth Derivative of Standard Functions
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