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We have to find the second derivative of arc tan (x)
[arc tan (x)]'' = [1/(1 + x^2)]'
[1/(1 + x^2)]' = [(1 + x^2)^-1]'
=> -1*(1 + x^2)^-2*2x
=> -2x/(1 + x^2)^2
The second integral of arc tan x = -2x/(1 + x^2)^2
For the beginning, we'll calculate the first derivative of f(x):
Now, we'll calculate f"(x) of the expression arctan x, or we'll calculate the first derivative of f'(x).
f"(x) = [f'(x)]'
Since f'(x) is a ratio, we'll apply the quotient rule:
f"(x) = [1'*(1+x^2)-1*(1+x^2)']/(1+x^2)^2
We'll put 1' = 0 and we'll remove the brackets:
Because of the fact that denominator is always positive, then the numerator will influence the ratio.
We notice that because of the fact that numerator is negative over the interval [0,infinite) => f"(x)<0.
Therefore, the 2nd derivative of the function y=arctan x is y"= -2x/(1+x^2)^2 and the graph of the function is concave over the positive real set of numbers.
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