It is important to know the derivatives of the inverse circular functions! Let's derive the important expression %(tanl(m)) = 1:562 ' lfy = tan1(3) then \"Sm- Now by implicit differentiation with respect to a: : 1 = jggoano = 3:(sec2(y)) . So as a function of y: But now we recall the identity sec2 (y) = 1 + tan2 (3;). So as a function of a: d\": 1ltan2(y) .m- Note: the Maple syntax for sin2 (3;) is sin(x) \"2. QED