Finding the Derivatives of tan x, cot x, sec x, and csc x

You already know the derivatives of sin x and cos x. The other four trig functions are built from those two as fractions, so the quotient rule hands you their derivatives for free. Write tan x as sin x over cos x, apply the quotient rule, and the numerator simplifies using sin² x + cos² x = 1 until only sec² x is left. The same process on cos x over sin x gives cot x a derivative of -csc² x. Near x = π/2, cos x approaches 0, so tan x shoots up to infinity and its slope, sec² x, blows up too since sec x = 1/cos x is also exploding there.

Sec x and csc x are reciprocals of cos x and sin x, so their derivatives come out as sec x tan x and -csc x cot x. Notice the pattern: every 'co' function (cosine, cotangent, cosecant) picks up a negative sign in its derivative, while tan, sec, and sin do not. Memorize these four alongside sin x and cos x, because from here on they show up constantly inside the product, quotient, and chain rules.

This topic is where differentiation rules stop being separate tricks and start working together. Expect to combine these derivatives with the product rule, as in sec x tan x, or with the chain rule, as when csc² x requires an extra factor of -csc x cot x from differentiating the inside function. Getting these six derivatives automatic now makes implicit differentiation and related rates, both coming up soon, far less error prone.