# What is the derivative of #x/(1+x^2)#?

We will use the following Quotient Rule for the Derivative :-

Hence,

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To find the derivative of ( \frac{x}{1+x^2} ), you can use the quotient rule.

The quotient rule states that if you have a function in the form ( \frac{u(x)}{v(x)} ), then its derivative is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

Now, applying the quotient rule to the given function:

Let ( u(x) = x ) and ( v(x) = 1 + x^2 ).

Then ( u'(x) = 1 ) and ( v'(x) = 2x ).

Plug these values into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{x}{1+x^2} \right) = \frac{(1)(1+x^2) - (x)(2x)}{(1+x^2)^2} ]

Simplify the expression:

[ \frac{1 + x^2 - 2x^2}{(1+x^2)^2} = \frac{1 - x^2}{(1+x^2)^2} ]

So, the derivative of ( \frac{x}{1+x^2} ) is ( \frac{1 - x^2}{(1+x^2)^2} ).

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