Calculus/Derivatives of Trigonometric Functions

Sine, cosine, tangent, cosecant, secant, cotangent. These are functions that crop up continuously in mathematics and engineering and have a lot of practical applications. They also appear in more advanced mathematics, particularly when dealing with things such as line integrals with complex numbers and alternate representations of space like spherical and cylindrical coordinate systems.

We use the definition of the derivative, i.e.,
 * $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ ,

to work these first two out.

Let us find the derivative of sin(x), using the above definition.

Now for the case of cos(x).

Therefore we have established

To find the derivative of the tangent, we just remember that:

$$\tan(x)=\frac{\sin(x)}{\cos(x)}$$

which is a quotient. Applying the quotient rule, we get:

$$\frac{d}{dx}\tan(x)=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}$$

Then, remembering that $$\cos^2(x)+\sin^2(x)=1$$, we simplify:

For secants, we again apply the quotient rule.


 * $$\sec(x)=\frac{1}{\cos(x)}$$
 * $$\begin{align}\frac{d}{dx}\sec(x)&=\frac{d}{dx}\frac{1}{\cos(x)}\\

&=\frac{\cos(x)\frac{d 1}{dx}-1\frac{d\cos(x)}{dx}}{\cos(x)^2}\\ &=\frac{\cos(x)0-1(-\sin(x))}{\cos(x)^2} \end{align}$$

Leaving us with:


 * $$\frac{d}{dx}\sec(x)=\frac{\sin(x)}{\cos^2(x)}$$

Simplifying, we get:

Using the same procedure on cosecants:
 * $$\csc(x)=\frac{1}{\sin(x)}$$

We get:

Using the same procedure for the cotangent that we used for the tangent, we get: