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.

$\frac{\mathrm{d}}{\mathrm{d}x}\sin u=\cos u$

$\frac{\mathrm{d}}{\mathrm{d}x}\cos u=-\sin u$

$\frac{\mathrm{d}}{\mathrm{d}x}\tan u={\sec }^{2}u$

$\frac{\mathrm{d}}{\mathrm{d}x}\cot u=-{\csc }^{2}u$

$\frac{\mathrm{d}}{\mathrm{d}x}\sec u=\sec u\tan u$

$\frac{\mathrm{d}}{\mathrm{d}x}\csc u=-\csc u\cot u$

Proofs

Proof for the derivative of $\sin u$

Let $y=\sin u$,

then

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

therefore

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

In Trigonometry,

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

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

we have

$\mathrm{\Delta }y=2\cos (u+\frac{\mathrm{\Delta }u}{2})sin\frac{\mathrm{\Delta }u}{2}$

Hence

$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\cos (u+\frac{\mathrm{\Delta }u}{2})\frac{\sin \frac{\mathrm{\Delta }u}{2}}{\frac{\mathrm{\Delta }u}{2}}$

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

$\frac{\sin \frac{\mathrm{\Delta }u}{2}}{\frac{\mathrm{\Delta }u}{2}}$

Hence

$\frac{\mathrm{d}y}{\mathrm{d}x}=\cos u$

Proof for $\frac{\mathrm{d}}{\mathrm{d}x}\cos u$

$\frac{\mathrm{d}}{\mathrm{d}x}\cos u=\sin u(-\frac{\mathrm{d}u}{\mathrm{d}x})=-\sin u\frac{\mathrm{d}u}{\mathrm{d}x}$

Proof for $\frac{\mathrm{d}}{\mathrm{d}x}\tan u$

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

$\frac{\mathrm{d}}{\mathrm{d}x}\tan u=\frac{\cos u\frac{\mathrm{d}}{\mathrm{d}x}\sin u-\sin u\frac{\mathrm{d}}{\mathrm{d}x}\cos u}{{\cos }^{2}u}$

$=\frac{{\cos }^{2}u\frac{\mathrm{d}u}{\mathrm{d}x}+{\sin }^{2}u\frac{\mathrm{d}u}{\mathrm{d}x}}{{\cos }^{2}u}=\frac{\mathrm{d}u}{{\cos }^{2}u}$

$={\sec }^{2}u\frac{\mathrm{d}u}{\mathrm{d}x}$

Proof for $\frac{\mathrm{d}}{\mathrm{d}x}\sec u$

$\frac{\mathrm{d}}{\mathrm{d}x}\tan u=\frac{\cos u\frac{\mathrm{d}}{dx}\sin u-\sin u\frac{\mathrm{d}}{\mathrm{d}x}\cos u}{{\cos }^{2}u}$

$=\frac{{\cos }^2\frac{\mathrm{d}u}{\mathrm{d}x}+{\sin }^{2}u\frac{\mathrm{d}u}{\mathrm{d}x}}{{\cos }^{2}u}=\frac{\frac{\mathrm{d}u}{\mathrm{d}x}}{{\cos }^{2}u}$

$={\sec }^{2}u\frac{\mathrm{d}u}{\mathrm{d}x}$