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

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Answer 1 · 2017-05-08T18:48:52

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

Knowing that, #sin2theta=(2tantheta)/(1+tan^2theta),# we put,

#x=tantheta# in #u#, and find that,

#u=tantheta/(1+tan^2theta)=1/2sin2theta.#

Note that, as the Range of #tan# fun. is #RR,# we can put #x=tantheta.#

Thus, #u# is a fun. of #theta#, and, #theta# of #x.#

Hence, by the Chain Rule, we have,

#(du)/dx=(du)/(d(theta))*(d(theta))/dx..........(star).#

Now, #u=1/2sin2theta rArr (du)/(d(theta))=1/2(cos2theta)(2), i.e.,#

#(du)/(d(theta))=cos2theta.................(1).#

#x=tantheta rArr theta=arc tanx rArr (d(theta))/dx=1/(1+x^2)....(2).#

Therefore, by #(star),(1) and (2),# we have,

#(du)/dx=(cos2theta)/(1+x^2)," and, finally, since,"#

#cos2theta=(1-tan^2theta)/(1+tan^2theta)=(1-x^2)/(1+x^2),#

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

Enjoy Maths.!