How to differentiate tan(x)?

tanx = sinx/cosx. Express this as: sinx*(cosx)^-1. Remembering the product rule: "y = f(x)g(x), dy/dx = f'(x)g(x) + f(x)g'(x)". sinx differentiates to cosx and cosx differentiates to -sinx. Also, remembering the chain rule, (cosx)^-1 differentiates to -sinx*(cosx)^-2. So tanx differentiates to cosx*(cosx)^-1 + sinxsinx(cosx)^-2. Which simplifies to 1 + (tanx)^2. From the identity (sinx)^2 + (cosx)^2 = 1, it can be seen that dividing through by (cosx)^2 gives (tanx)^2 + 1 = (cosx)^-2. (secx)^2 equals (cosx)^-2. Hence (tanx)^2 + 1 = (secx)^2. Therefore differentiating tanx gives (secx)^2.