How do you find the derivative of #x/(1+x^2)#?

2 Answers
F. Javier B.
Jun 3, 2018

Use the rule #(f/g)´=(f´·g-f·g´)/g^2#

Explanation:

If #y=x/(1+x^2)#

Then

#y´=(1·(1+x^2)-x·2x)/(1+x^2)^2=(1-x^2)/(1+x^2)^2#

Where #f=x# and #g=1+x^2#

Jim G.
Jun 3, 2018

#(1-x^2)/(1+x^2)^2#

Explanation:

#"differentiate using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=xrArrg'(x)=1#

#h(x)=1+x^2rArrh'(x)=2x#

#d/dx(x/(1+x^2))#

#=(1+x^2-2x^2)/(1+x^2)^2=(1-x^2)/(1+x^2)^2#

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