Derivatives of trigonometric functions

You already know how to take a derivative of polynomial. In this topic, you will learn about a derivative of a trigonometric function: sine, cosine, tangent and cotangent.

Formulas of derivatives of important trigonometric functions

As you remember, a derivative of a function $f(x)$ could be found with the following formula:

$$f'(x) = \lim\limits_{\triangle x \to 0} \dfrac{f(x + \triangle x) - f(x)}{\triangle x}$$Let's find a derivative of sine of $x$. Using our formula,

$$(\sin x)' = \lim\limits_{\triangle x\to 0} \dfrac{\sin(x + \triangle x) - \sin x}{\triangle x}$$Given the trigonometric identity$$\sin \alpha - \sin \beta = 2 \sin \dfrac{\alpha - \beta}{2}\cos \dfrac{\alpha + \beta}{2}$$

and the first remarkable limit, we get:

$$(\sin x)' = \lim\limits_{\triangle x\to 0} \dfrac{2\sin\dfrac{\triangle x}{2}\cos \left( x + \dfrac{\triangle x}{2}\right)}{\triangle x} = \cos x$$So, a derivative of $\sin x$ is equal to $\cos x$.

Similarly, we can get that

$$(\cos x)' = -\sin x$$With this in mind, we derive a formula for the derivative of tangent using the quotient rule:

$$(\tan x)' = \left( \dfrac{\sin x}{\cos x}\right)' = \dfrac{(\sin x)' \cdot \cos x - \sin x \cdot (\cos x)'}{\cos^2 x} = \\ \ \\ =\dfrac{\cos x \cdot \cos x - \sin x\cdot (-\sin x)}{\cos^2x} = \dfrac{\cos^2x + \sin^2x}{\cos^2 x} = \dfrac{1}{\cos^2 x}$$So, the derivative of tangent of $x$ is equal to $\dfrac{1}{\cos^2 x}$.

Similarly, you can derive the formula for the cotangent of $x$:

$$(\cot x)' = - \dfrac{1}{\sin^2 x}$$

Examples

Let's find a derivative of $f(x) = \tan x + \cos x$. Using the sum rule, we get:

$$f'(x) = (\tan x + \cos x)' = (\tan x)' + (\cos x)' = \dfrac{1}{\cos^2 x} - \sin x$$Consider an another example. Let's find a derivative of $f(x) = \dfrac{x}{\tan x}$. Using the quotient rule, we get:

$$f'(x) = \left(\dfrac{x}{\tan x}\right)' = \dfrac{x'\ \cdot \tan x - x \cdot (\tan x)'}{\tan^ 2x} = \dfrac{\tan x - \dfrac{x}{\cos^2 x}}{\tan^2 x} = \\ \ \\ = \dfrac{1}{\tan x} - \dfrac{x}{\cos^2 x \cdot \tan^2 x} = \dfrac{1}{\tan x} - \dfrac{x}{\sin^2 x}$$

Conclusion

In this topic, you have learned about derivatives of important trigonometric functions:

$(\sin x)'=\cos x$

$(\cos x)'=-\sin x$

$(\tan x)' = \dfrac{1}{\cos^2 x}$

$(\cot x)' = - \dfrac{1}{\sin^2 x}$