What is the n^(th) derivative of y = x^(2n) ?
2 Answers
Let's have a look.
Explanation:
Let
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Following the sequences...
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Hope it Helps:)
y^((n)) = ((2n)!) / (n!) \ x^n
Explanation:
y = x^(2n)
We can calculate the first few derivatives directly:
y^((1)) = (2n)x^(2n-1)
y^((2)) = (2n)(2n-1)x^(2n-2)
y^((3)) = (2n)(2n-1)(2n-2)x^(2n-3)
From which we may conclude:
y^((n)) = (2n)(2n-1)(2n-2)...(2n-(n+1))x^(2n-n)
\ \ \ \ \ \ \ = (2n)(2n-1)(2n-2)...(2n-n-1)x^(2n-n)
\ \ \ \ \ \ \ = (2n)(2n-1)(2n-2)...(n-1)x^(2n-n)
\ \ \ \ \ \ \ = ((2n)(2n-1)(2n-2)...(n-1)n(n-1)...1x^(2n-n)) / (n(n-1)...1)
\ \ \ \ \ \ \ = ((2n)!) / (n!) \ x^n