What is the n^(th) derivative of y = x^(2n) ?

2 Answers
Jan 16, 2018

Let's have a look.

Explanation:

Let y be a function of x, such that,

y=f(x)

:.color(red)(y=x^(2n))

Now, let y_n be the n^(th) derivative of the function y.

:.y_1=2nx^(2n-1)

:.y_2=2n(2n-1)nx^(2n-2)

:.y_3=2n(2n-1)(2n-2)nx^(2n-3)
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Following the sequences...
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:.color(red)(y_n=[2n(2n-1)....(2n-n+1)]nx^(2n-n))

:.color(red)(y_n=nx^n[2n(2n-1)....(n+1)])

Hope it Helps:)

Jan 17, 2018

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