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 #