f(x)=1/(x^2+1)

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(1/(1+x^2))
The derivative of f(x) is f'(x):
f'(x) = 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:
f'(x) = -(d/dx(1+x^2))/((x^2+1)^2)
Differentiate the sum term by term:
f'(x) = -1/(1+x^2)^2 d/dx(1)+d/dx(x^2)
The derivative of 1 is zero:
f'(x) = -(d/dx(x^2)+0)/(1+x^2)^2
Simplify the expression:
f'(x) = -(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:
f'(x) = -1/(1+x^2)^2 2 x
Expand the left hand side:
Answer: |f'(x) = -(2 x)/(1+x^2)^2

Firstly - convert the fraction into a linear equation;

$1/(x^2+1)$ goes to $1\times{(x^2+1)}^{-1}$

Now you can just use the chain rule - or whatever you have been taught

let $u=x^2+1$

therefore $f(x) = 1+ {(u)}^{-1}$

therefore$\frac{dy}{dx}(1+{(u)}^{-1})=-1{(u)}^{-2}$

thats part one \o/

Part two;

if $u=x^2+1$

therefore $\frac{dy}{dx}(x^2+1)=2x^1+0$

Combine part two and part one;

$(2x)(-1){(u)}^{-2}$

sub in for $u$

$(2x)(-1){(x^2+1)}^{-2}$

simplifiy brackets

$-2x{(x^2+1)}^{-2}$

turn into a fraction

$\frac{-2x}{(x^2+1){}^{2}}$

ta dah!