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

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Answer 1 · 2016-12-31T16:57:08

#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)#.