Question #bf4da

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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:

$v'=1-v^2$

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