High School Calculus/Derivatives of Trigonometric Functions

Formulas for Differentiation of Trigonometric Functions
In the following formulas the angle u is supposed to be expressed in circular measure.

$${\operatorname {d} \over \operatorname {d}x}\sin u = \cos u$$

$${\operatorname {d} \over \operatorname {d}x} \cos u= -\sin u$$

$${\operatorname {d} \over \operatorname {d}x} \tan u= \sec ^2 u$$

$${\operatorname {d} \over \operatorname {d}x} \cot u= -\csc ^2 u$$

$${\operatorname {d} \over \operatorname {d}x} \sec u = \sec u \tan u$$

$${\operatorname {d} \over \operatorname {d}x} \csc u = -\csc u \cot u$$

Proofs
Proof for the derivative of $$\sin u$$

Let $$y = \sin u$$,

then $$y \prime = \sin (u + \Delta u)$$;

therefore $$\Delta y= \sin (u + \Delta u) - \sin u$$;

In Trigonometry, $$\sin A - \sin B = 2\sin {1 \over 2}(A - B)\cos {1 \over 2}(A + B)$$

If we substitute $$A = u + \Delta u$$ and $$B = u$$,

we have $$\Delta y = 2\cos \left(u + { \Delta u \over 2}\right) \ sin {\Delta u \over 2}$$

Hence $${\Delta y \over \Delta x} = \cos \left(u + {\Delta u \over 2}\right){\sin {\Delta u \over 2} \over {\Delta u \over 2}}$$

When $$\Delta x$$ approaches zero, $$\Delta u$$ likewise approaches zero, and as $$\Delta u$$ is in circular measure, the limit of

$${\sin {\Delta u \over 2} \over {\Delta u \over 2}}$$

Hence $${\operatorname {d}y \over \operatorname {d}x} = \cos u$$

Proof for $${\operatorname {d} \over \operatorname {d}x}\cos u$$

$${\operatorname {d} \over \operatorname {d}x}\cos u=\sin u \left(-{\operatorname {d}u \over \operatorname {d}x}\right)=-\sin u {\operatorname {d}u \over \operatorname {d}x}$$

Proof for $${\operatorname {d} \over \operatorname {d}x} \tan u$$

Since $$\tan u = {\sin u \over \cos u}$$

$${\operatorname {d} \over \operatorname {d}x}\tan u = {\cos u {\operatorname {d} \over \operatorname {d}x}\sin u - \sin u {\operatorname {d} \over \operatorname {d}x}\cos u \over \cos ^2 u}$$

$$={\cos ^2 u {\operatorname {d}u \over \operatorname {d}x} + \sin ^2 u {\operatorname {d}u \over \operatorname {d}x} \over \cos ^2 u} = {\operatorname {d}u \over \cos ^2 u}$$

$$=\sec ^2 u {\operatorname {d}u \over \operatorname {d}x}$$

Proof for $${\operatorname {d} \over \operatorname {d}x}\sec u$$

$${\operatorname {d} \over \operatorname {d}x}\tan u = {\cos u {\operatorname {d} \over {d}x}\sin u - \sin u {\operatorname {d} \over \operatorname {d}x}\cos u \over \cos ^2 u}$$

$$={\cos ^2 {\operatorname {d}u \over \operatorname {d}x} + \sin ^2 u {\operatorname {d}u \over \operatorname {d}x} \over \cos ^2 u} = {{\operatorname {d}u \over \operatorname {d}x} \over \cos ^2 u}$$

$$=\sec ^2 u{\operatorname {d}u \over \operatorname {d}x}$$