find the derivative of tan(2x)

Answer

`[[cos theta cos phi cos(theta ~phi), cos theta sin phi cos (theta~phi) ],[sintheta cos phi cos (theta~phi), sin theta sin phi cos (theta~phi) ]]`

Explanation

`AB = [[cos^(2)theta,costhetasintheta],[costhetasintheta,sin^(2)theta]][[cos^(2)phi,cosphisinphi],[cosphisinphi,sin^(2)phi]]`
` = {:[[cos^(2)theta cos^(2)phi+cos theta cos phi sintheta sinphi],[cos^(2)phi cos theta sin theta+sin^(2)theta sin phi cos phi]:}`
` {:[cos^(2)theta cosphisin phi+sin^(2) phi sinthetacos theta ],[cos theta cos phi sin thetasin phi+sin^(2)theta sin^(2) phi ]]`
` {:[[costheta cosphi (costheta cosphi + sintheta sin phi ) ],[sin theta cos phi (cos theta cos phi + sin theta sin phi)]:}`
` {:[cosphi sinphi (costheta cosphi + sintheta sin phi ) ],[sin theta sin phi (cos theta cos phi + sin theta sin phi)]]`
`[[cos theta cos phi cos (theta-phi),cos theta sin phi cos(theta-phi)],[sin theta cos phi (theta - phi),sin theta sin phi cos (theta-phi)]]`
Clearly, AB is the zero matrix, if cos `theta~ phi) = 0` i.e., `theta - phi` is an
odd multiple of `pi/2`.