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

Question

Christopher Agresta · Accepted Answer

#dy/dx=(1-x^2)/(1+x^2)#.

Let #y=x/(1+x^2)#. We will use the following Quotient Rule for the Derivative :- #y=(u(x))/(v(x)) rArr dy/dx={v(x)u'(x)-u(x)v'(x)}/(v(x))^2# Hence, #dy/dx={(1+x^2)(x)'-x(1+x^2)'}/(1+x^2)^2# #=[(1+x^2)(1)-(x){1'+(x^2)'}]/(1+x^2)^2# #={(1+x^2)-x(0+2x)}/(1+x^2)^2# #=(1+x^2-2x^2)/(1+x^2)^2# #rArr dy/dx=(1-x^2)/(1+x^2)#.

Luke Alvarez · Answer

undefined 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} $.