This is equivalent, yes?pic.twitter.com/TSvUw1IzcT
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A proof y = tanx; z = -cotx => y' = (secx)^2; z' = (cscx)^2; clearly (secx)^2 + (cscx)^2 = (secx)^2(cscx)^2 (e^w)' = (e^w)w' with w = y or w = z First member is (e^(y+z))(y' + z') = (y'e^y)(z'e^z) = (e^y)'(e^z)' QED.
Thanks. Twitter will use this to make your timeline better. UndoUndo
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