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

Logo

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} =$


See also

Interested in discovering more concerning by-products? You can have a look at these web pages:



Source link .