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

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

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

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