What is the #n^(th)# derivative of # y = x^(2n) #?
2 Answers
Let's have a look.
Explanation:
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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 #