## Formulas for the by-products of trigonometric functions

### Derivative of the sine function

The by-product of the basic sine feature is:

\$ latex transgression ^ {prime} (x)= cos( x)\$

To obtain sine features of the kind \$latex transgression( nx)\$, we utilize the chain policy with \$latex y= transgression( u)\$ and also \$latex u= nx\$.

Likewise, to obtain features of the kind \$latex transgression ^ n( x)=( transgression( x)) ^ n\$, we utilize the chain policy with \$latex y= u ^ n\$ and also \$latex u= transgression( x)\$.

### By-product of the cosine function

The by-product of the basic cosine feature is:

\$ latex cos ^ {prime} (x)=- transgression( x)\$

If we have features of the kind \$latex cos( nx)\$, we can utilize the chain policy with \$latex y= cos( u)\$ and also \$latex u= nx\$.

For features of the kind \$latex cos ^ n( x)=( cos( x)) ^ n\$, we utilize the chain policy with \$latex y= u ^ n\$ and also \$latex u= cos( x)\$.

### By-product of the tangent function

The by-product of the basic tangent feature is:

\$ latex tan ^ {prime} (x)= sec ^ 2( x)\$

For features of the kind \$latex tan( nx)\$, we utilize the chain policy with \$latex y= tan( u)\$ and also \$latex u= nx\$.

For features of the kind \$latex tan ^ n( x)=( tan( x)) ^ n\$, we utilize the chain policy with \$latex y= u ^ n\$ and also \$latex u= tan( x)\$.

### By-product of the cosecant function

The by-product of the basic cosecant feature is:

\$ latex cosec ^ {prime} (x)=- cosec( x) cot( x)\$

The cosecant features of the kind \$latex cosec( nx)\$, can be acquired with the chain policy by utilizing \$latex y= cosec( u)\$ and also \$latex u= nx\$.

Likewise, features of the kind \$latex cosec ^ n( x)=( cosec( x)) ^ n\$, are acquired with the chain policy with \$latex y= u ^ n\$ and also \$latex u= cosec( x)\$.

### By-product of the secant function

The by-product of the basic secant feature is:

\$ latex sec ^ {prime} (x)= sec( x) tan( x)\$

Secant features of the kind \$latex sec( nx)\$ are acquired utilizing the chain policy with \$latex y= sec( u)\$ and also \$latex u= nx\$.

Likewise, features of the kind \$latex sec ^ n( x)=( sec( x)) ^ n\$ are acquired utilizing the chain policy with \$latex y= u ^ n\$ and also \$latex u= sec( x)\$.

### By-product of the cotangent function

The by-product of the basic cotangent feature is:

\$ latex cot ^ {prime} (x)=- cosec ^ 2( x)\$

To obtain cotangent features of the kind \$latex cot( nx)\$, we use the chain policy with \$latex y= cot( u)\$ and also \$latex u= nx\$.

Features of the kind \$latex cot ^ n( x)=( transgression( x)) ^ n\$, are additionally acquired utilizing the chain policy with \$latex y= u ^ n\$ and also \$latex u= cot( x)\$.

## By-products of trigonometric features– Instances with answers

### EXAMPLE 1

Encuentra la derivada de \$latex y= transgression( 5x)\$.

We can utilize the chain policy with \$latex u= 5x\$. After that, we have:

\$ latex y= transgression( u)|| \$ and also \$latex|| u= 5x\$

Their by-products are:

\$ latex dfrac {dy} {du} =cos( u)|| \$ and also \$latex|| dfrac {du} {dx} =5\$

Now, we use the chain policy:

\$\$ dfrac {dy} {dx} =dfrac {dy} {du} dfrac {du} {dx} =cos( u) times 5\$\$

\$\$ dfrac {dy} {dx} =5cos (5x)\$\$

Generally, we create in the list below means to locate the solution quicker:

\$\$ dfrac {dy} {dx} =cos (5x) times (5x) ^ {prime} =5cos( 5x)\$\$

### EXAMPLE 2

Find the by-product of \$latex y= cos ^ 2( x)\$.

We can begin by composing as \$latex (cos( x)) ^ 2\$. After that, we have:

\$ latex y= u ^ 2|| \$ and also \$latex|| u= cos( x)\$

Their by-products are:

\$ latex dfrac {dy} {du} =2u|| \$ and also \$latex|| dfrac {du} {dx} =- transgression( x)\$

Using the chain policy, we have:

\$\$ dfrac {dy} {dx} =dfrac {dy} {du} dfrac {du} {dx} =2u(- transgression( x))\$\$

\$\$ dfrac {dy} {dx} =-2 cos( x) transgression( x)\$\$

Generally, we create in the list below means to locate the solution quicker:

\$\$ dfrac {dy} {dx} =2cos (x) times (cos( x)) ^ {prime} =-2 cos( x) transgression( x)\$\$

### EXAMPLE 3

Find the by-products of the complying with features:

a) \$latex y= transgression( x ^ 2 +2)|||\$ b) \$latex y= cos( sqrt {x} )\$

a) When \$latex y= transgression( x ^ 2 +2)\$, we have:

\$\$ dfrac {dy} {dx} =cos (x ^ 2 +2) times (x ^ 2 +2) ^ {prime} \$\$

\$\$ dfrac {dy} {dx} =2xcos (x ^ 2 +2)\$\$

b) When \$latex y= cos( sqrt {x} )\$, we have:

\$\$ dfrac {dy} {dx} =- transgression( sqrt {x} )times (sqrt {x} )^ {prime} \$\$

\$\$ dfrac {dy} {dx} =- frac {1} {2sqrt {x}} transgression( sqrt {x} )\$\$

### EXAMPLE 4

What are the by-products of the complying with features?

a) \$latex y= transgression ^ 4( x)|||\$ b) \$latex y= cos ^ 3( 2x)\$

a) When \$latex y= transgression ^ 4( x)\$, we have:

\$\$ dfrac {dy} {dx} =4sin ^ 3 (x) times (transgression( x)) ^ {prime} \$\$

\$\$ dfrac {dy} {dx} =4sin ^ 3 (x) cos (x)\$\$

b) When \$latex y= cos ^ 3( 2x)=( cos( 2x)) ^ 3\$, we have:

\$\$ dfrac {dy} {dx} =3( cos( 2x)) ^ 2times (cos( 2x)) ^ {prime} \$\$

\$\$= 3( cos( 2x)) ^ 2times (-2 transgression( 2x)) ^ {prime} \$\$

\$\$ dfrac {dy} {dx} =-6 cos ^ 2( 2x) transgression( 2x)\$\$

### EXAMPLE 5

Derives the list below features:

a) \$latex y= tan( 3x)|||\$ b) \$latex y= 4cosec ^ 2( x)\$

a) When \$latex y= tan( 3x)\$, we have:

\$\$ dfrac {dy} {dx} =sec ^ 2( 3x) times (3x) ^ {prime} \$\$

\$\$ dfrac {dy} {dx} =3sec ^ 2 (3x)\$\$

b) When \$latex y= 4cosec ^ 2( x)\$, we have:

\$\$ dfrac {dy} {dx} =8cosec( x) times (cosec( x)) ^ {prime} \$\$

\$\$= 8cosec( x)(- cosec( x) cot( x))\$\$

\$\$ dfrac {dy} {dx} =-8 cosec ^ 2( x) cot( x)\$\$

## Derivatives of trigonometric features– Technique problems You have actually finished the test!

#### What is the by-product of \$latex y= cosec( x-1)\$?

Create the solution in the input box.

\$ latex frac {dy} {dx} =\$