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Answer 1 · 2017-11-14T04:46:32

The answer is :

#u'-v'=u^2+v^2#

Using the quotient rule for differentiation:

#u'=((cosx-sinx)(cosx-sinx)-(cosx+sinx)(-cosx-sinx))/(cosx-sinx)^2#

#u'=((cosx-sinx)^2+(cosx+sinx)^2)/(cosx-sinx)^2#

#u'=1+((cosx+sinx)^2)/(cosx-sinx)^2=1+u^2#

From

#u=1/v#

we get

#v=1/u=(cosx-sinx)/(cosx+sinx)#

Using the quotient rule again we arrive at:

#v'=1-v^2#

#u'-v'=1+u^2-(1-v^2)=1+u^2-1+v^2=u^2+v^2#