The calculation is in the picture. Please help.
http://farm7.static.flickr.com/6106/6271…
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f(x)= ln (2x^2 + x-1) - ln(x+1)
f(x)=ln(2x-1)
first derivative = 2*(2x-1)^(-1)
second derivative = 2*(-1)*(2x-1)^(-2) * (2)
third derivative = 2*(-1)*(-2)*(2x-1)^(-3) * (2)
nth derivative=2 *[(n-1)!] *(2x-1)^(-n) * 2 ........ is something wrong here?
or is it
nth derivative=2 *[(n-1)!] *(2x-1)^(-n) * 2^(-1)=[(n-1)!] *(2x-1)^(-n)..............or is this wrong?
The correct answer to the equation is -2^(-49) * 97!
http://farm7.static.flickr.com/6106/6271…
from http://www.flickr.com/
f(x)= ln (2x^2 + x-1) - ln(x+1)
f(x)=ln(2x-1)
first derivative = 2*(2x-1)^(-1)
second derivative = 2*(-1)*(2x-1)^(-2) * (2)
third derivative = 2*(-1)*(-2)*(2x-1)^(-3) * (2)
nth derivative=2 *[(n-1)!] *(2x-1)^(-n) * 2 ........ is something wrong here?
or is it
nth derivative=2 *[(n-1)!] *(2x-1)^(-n) * 2^(-1)=[(n-1)!] *(2x-1)^(-n)..............or is this wrong?
The correct answer to the equation is -2^(-49) * 97!
-
With a cherry on top, I want to know where you went wrong too! I went through your process and came up with the same answer, the only feasible way of getting -2^49 is being have (2^.5)^-104. According to your formula, that would only be possible on the 104th derivative. Are you sure that's is the right answer?
Edit: 2 *[(n-1)!] *(2x-1)^(-n) * 2^(-1) makes more sense, 2^-1 and 2 always cancel out no matter what. Now you're left with 97!X2^-49!!!
But how would we be able to get 1/2 out of 2x-1....
Edit: 2 *[(n-1)!] *(2x-1)^(-n) * 2^(-1) makes more sense, 2^-1 and 2 always cancel out no matter what. Now you're left with 97!X2^-49!!!
But how would we be able to get 1/2 out of 2x-1....