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.

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

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

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

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

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

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

Proofs

Proof for the derivative of ${\displaystyle \sin u}$

Let ${\displaystyle y=\sin u}$,

then

${\displaystyle y\prime =\sin(u+\Delta u)}$;

therefore

${\displaystyle \Delta y=\sin(u+\Delta u)-\sin u}$;

In Trigonometry,

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

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

we have

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

Hence

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

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

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

Hence

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

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

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

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

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

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

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

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

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

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

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

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