Post a Comment

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

Got questions relating to this, Feel free to ask our telegram community

Help us grow our readership by sharing this post

Related Posts

Post a Comment

Subscribe Our Newsletter