How can I find the derivative of y = tan(x)?

Let's first recall the definition of tan(x) = sin(x)/cos(x). Hence, y' = d/dx( sin(x)/cos(x) ). Recall quotient rule for differentiation: ( f(x)/g(x) )' = ( f '(x)^2*g(x) - f(x)*g'(x)^2 ) / g(x)^2 and that d/dx( sin(x) ) = cos(x), and d/dx( cos(x) ) = - sin(x). Then, y' = ( cos(x)cos(x) - sin(x)(-sin(x)) )/cos^2(x) => y' = ( cos^2(x) + sin^2(x) ) /cos^2(x). Recalling that cos^2(x) + sin^2(x) = 1 by the Pythagorean trigonometric identity, then y' = 1 / cos^2(x). Since sec(x) = 1/cos(x), we can state our result as: y' = d/dx( tan(x) ) = sec^2(x).