What would the Derivative of (1/(1+x^2)) be? why?

Answer 1

y=1/(1+x^2)= (1+x^2)^(-1)
dy/dx= (-1)(1+x^2)^(-2) (2x)
=-2x /(1+x^2)^2
if you differentiate x^n = n x^(n-1) (we have used this principle). Note the 2x.

Answer 2

-2x/(1 + x²)²
Think of it as (1 + x²)^(-1) and use the Chain Rule.

Doug

Answer 3

let 1+x^2 = t.

1.Differentiate with respect to 't'.
d(y)/d(t)= d(t^ -1) = -1*(t^-2) [since d(x^n) = n*(x^(n-1)) ]
= -1/(t^2)

2. d(t)/d(x)= 2x.

3.Multiply by differentiation of 't' with respect to 'x'
i.e. d(y)/d(t) * d(t)/d(x) to get d(y)/d(x):
-1/(t^-2) * 2x
=-2x/t^2
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