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the derivative of 1/(1+x^2)

Guest Mar 16, 2017
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Possible derivation:
d/dx(1/(1 + x^2))
Using the chain rule, d/dx(1/(x^2 + 1)) = d/( du)1/u 0, where u = x^2 + 1 and ( d)/( du)(1/u) = -1/u^2:
 = -(d/dx(1 + x^2))/(1 + x^2)^2
Differentiate the sum term by term:
 = -1/(1 + x^2)^2 d/dx(1) + d/dx(x^2)
The derivative of 1 is zero:
 = -(d/dx(x^2) + 0)/(1 + x^2)^2
Simplify the expression:
 = -(d/dx(x^2))/(1 + x^2)^2
Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2: d/dx(x^2) = 2 x:
Answer: |= -2x / (1 + x^2)^2

Guest Mar 16, 2017

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