I would really like some Calculus help.

Reply Wed 15 Oct, 2014 01:48 am

I need to find an expression for the nth derivative of this function. Finding the first few derivatives isn't a problem, and I can see a relationship between them, but I can't figure out how to relate that to n. Thanks in advance!
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Reply Wed 15 Oct, 2014 08:04 am
f(x) = 1/(1+2x) = (1+2x) ^ (-1)
f'(x) = (-1) (2) (1+2x) ^ (-2)
f''(x) = (-1)(-2) (2)^2 (1+2x) ^ (-3)

Do you see the pattern here?
The first term is the sign going from negative to positive to negative, etc. That means (-1)^n.
The second term is going to be 1*2*3*...*n coming from that negative exponent getting more negative each time = n!
The third term is that 2^n term. It comes from the derivative of (1+2x) repeated over and over.
Finally the fourth term is (1+2x) ^ (-n+1) because it started at one not zero.

f^n(x) = (-1)^n (2^n) (n!) (1+2n)^-(n+1)
Reply Wed 15 Oct, 2014 11:37 am
Thank you so much! I had the (-1)^n and (1+2x)^-(n+1) parts right but I didn't even think of using a factorial for the rest. Much appreciated.
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