A trigonometric function is one with sine, cosine, tangent, secant, cosecant, cotangent.

In differentiating trigonometric function, Here are some things you need to know:

$$\frac{d}{dx}(\sin x)=\cos x$$

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

$$\frac{d}{dx}(\tan x)=\sec^2 x$$

$$\frac{d}{dx}(\sec x)=\sec x \tan x$$

$$\frac{d}{dx}(\csc x)=-\csc x \cot x$$

$$\frac{d}{dx}(\cot x)=-\csc^2 x$$

Example 1

Differentiate $y=\sin x+ \cos x-\csc x$


$$\frac{dy}{dx}=\cos x+(-\sin x)-\csc x \cot x$$

$$\frac{dy}{dx}=\cos x-\sin x-\csc x \cot x$$

Example 2

Differentiate $y=\tan x+ \sec x$


$$\frac{dy}{dx}=\sec^2 x+\sec x \tan x$$

$$\frac{dy}{dx}=\sec x( \sec x+ \tan x)$$

Example 3

Differentiate the function $h(x)=2\sin x \tan x$


$2\sin x \tan x$ is the product of $2\sin x$ and $\tan x$, Hence, we use the product rule which states that:


Here, $f=2\sin x$, $f'=2\cos x$, $g=\tan x$, $g'=\sec² x$

Using the formula for the product rule

$h'(x)=(2\cos x \tan x)+(\sec^2 2\sin x)$

$h'(x)=2\cos x \tan x+2\sec^2 x \sin x$

$h'(x)=2(\cos x \tan x+\sec^2 \sin x)$

Example 4

If $y=\frac{\sec x}{1+\tan x}$, Evaluate $\frac{dy}{dx}$ 


This involves division, Hence we use the quotient rule which is expressed as:


Here $f=\sec x$, $f'=\sec x\tan x$, $g=1+\tan x$, $g'=\sec² x$

$\frac{dy}{dx}=\frac{(\sec x \tan x)(1+\tan x)-sec^2 x(sec x)}{(1+tan x)^2}$

$\frac{dy}{dx}=\frac{(\sec x \tan x)(1+\tan x)-sec^3 x}{(1+tan x)^2}$

$\frac{dy}{dx}=\frac{(\sec x \tan x)(1+\tan x)}{(1+\tan x)^2}-\frac{\sec^3 x}{(1+\tan x)^2}$

$\frac{dy}{dx}=\frac{\sec x\tan x}{1+\tan x}-\frac{\sec^3 x}{(1+\tan x)^2}$

For differentiation of trigonometric function, I recommend that you know the derivative of trigonometric expression given above. 

In this next post, You would be learning about the differentiation of composite trigonometric functions

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