**Trigonometric derivatives.**The derivative trig functions are a quite complex subtopic because not only must understand the concept of derivatives, but we must also understand the concept of trigonometry. On derivative trigonometric functions, there are some general provisions that have become the basic reference to solve the problems. However, sometimes we have to change the shape of the function

The trigonometry is given to another simpler form in order to approach that generalized pattern become a determination. In this section, we usually experience difficulties if we do not understand the concept of trigonometry. Therefore, to master the topic of turmo trig functions we must master the concept of trigonometry.

The derivation of a trigonometric function is a different trigonometric function. Here is presented the table of initial functions and derivatives of trigonometric functions that serve as the basic reference.

following trigonometric equation derivatives

1. f(x) = sin x → f '(x) = cos x

2. f(x) = cos x → f '(x) = −sin x

3. f(x) = cot x → f '(x) = −csc2 x

4. f(x) = tan x → f '(x) = sec2 x

5. f(x) = sec x → f '(x) = sec x . tan x

6. f(x) = csc x → f '(x) = −csc x . cot x

2. f(x) = cos x → f '(x) = −sin x

3. f(x) = cot x → f '(x) = −csc2 x

4. f(x) = tan x → f '(x) = sec2 x

5. f(x) = sec x → f '(x) = sec x . tan x

6. f(x) = csc x → f '(x) = −csc x . cot x

If u is a function that can be derived against x with u 'is a derivative u to x, then:

1. f(x) = sin u → f '(x) = cos u . u'

2. f(x) = cos u → f '(x) = −sin u . u'

3. f(x) = tan u → f '(x) = sec2u . u'

4. f(x) = cot u → f '(x) = −csc2 u . u'

5. f(x) = sec u → f '(x) = sec u tan u . u'

6. f(x) = csc u → f '(x) = −csc u cot u . u'

2. f(x) = cos u → f '(x) = −sin u . u'

3. f(x) = tan u → f '(x) = sec2u . u'

4. f(x) = cot u → f '(x) = −csc2 u . u'

5. f(x) = sec u → f '(x) = sec u tan u . u'

6. f(x) = csc u → f '(x) = −csc u cot u . u'

On this occasion we will share an example of a derivative of trigonometric functions.

**Problem No.1**

Find the derivative of y = sin 4x!

**Answer:**

for example :

u = 4x ⇒ u' = 4

y' = cos u . u'

y' = cos 4x . 4

y' = 4cos 4x

u = 4x ⇒ u' = 4

y' = cos u . u'

y' = cos 4x . 4

y' = 4cos 4x

**Problem No.2**

Find the derivative of y = cos x

^{2}

**Answer:**

For example :

u = x

y' = −sin u . u'

y' = −sin x

y' = −2x sin x

u = x

^{2}⇒ u' = 2xy' = −sin u . u'

y' = −sin x

^{2}. 2xy' = −2x sin x

^{2}**Problem No.3**

Find the derivative of y = tan (2x+1)

**Answer:**

For Example:

u = 2x + 1 ⇒ u' = 2

y' = se

y' = sec

y' = 2sec

u = 2x + 1 ⇒ u' = 2

y' = se

^{2}u . u'y' = sec

^{2}(2x+1) . 2y' = 2sec

^{2}(2x+1)**Problem No.4**

Define the first instance of y = sin 5x + cos 3x.

**Answer:**

y' =

dx dx

y' = 5 cos 5x − 3 sin 3x.

__dy__=__d (sin 5x + cos 3x)__dx dx

y' = 5 cos 5x − 3 sin 3x.

**Problem No.5**

Define the first instance of y = 2 sin 7x − 5 cos x

**Answer:**

y' =

dx dx

y' = 14 cos 7x − (-5 sin x)

y' = 12 cos 2x + 5 sin x

__dy__=__d (2 sin 7x − 5 cos x)__dx dx

y' = 14 cos 7x − (-5 sin x)

y' = 12 cos 2x + 5 sin x

**→ Read also :**Problem And Discussion Of Limit Of Trigonometric Functions

**Problem No.6 :**

If y = 4x

^{5}+ sin 3x + cos 8x, then specify the first instance

**Answer**

⇒ y' =

dx dx

⇒ y' = 20 x

__dy__=__d (____4x__^{5}+ sin 3x + cos 3x)dx dx

⇒ y' = 20 x

^{3}+ 3 cos 3x − 8 sin 8x.**Problem No.7 :**

If

*f (x)*= sin 3x cos 2x, then find f '(π / 3)

**Answer**

We can use the concept of derivative multiplication function. For example:

⇒ u(x) = sin 3x, then u'(x) = 3 cos 3x

⇒ v(x) = cos 2x, then v'(x) = -2 sin 2x

So the first derivative is:

f '(x) = dy = u' (x) .v (x) + u (x) .v '(x)

f '(x) = 3 cos 3x. (cos 2x) + 3 sin x. (-2 sin 2x)

f '(x) = 3 cos 3x. cos 2x − 6 sin x. sin 2x

f '(π⁄3 ) = 3 cos 3(π⁄3 ). cos 2(π⁄3 ) − 6 sin (π⁄3 ). sin 2(π⁄3 )

f '(π⁄3 ) = 3 cos π. cos (2π⁄3)- 6 sin (π⁄3 ). sin (2π⁄3 )

f '(π⁄3 ) =3 (-1). (-0,5) - 6(½√3) ½√3

f '(π⁄3 ) =1,5 - 6 ( 1⁄4 3)

f '(π⁄3 ) = 1,5 - 4,5

f '(π⁄3 ) = -3

⇒ u(x) = sin 3x, then u'(x) = 3 cos 3x

⇒ v(x) = cos 2x, then v'(x) = -2 sin 2x

So the first derivative is:

f '(x) = dy = u' (x) .v (x) + u (x) .v '(x)

f '(x) = 3 cos 3x. (cos 2x) + 3 sin x. (-2 sin 2x)

f '(x) = 3 cos 3x. cos 2x − 6 sin x. sin 2x

f '(π⁄3 ) = 3 cos 3(π⁄3 ). cos 2(π⁄3 ) − 6 sin (π⁄3 ). sin 2(π⁄3 )

f '(π⁄3 ) = 3 cos π. cos (2π⁄3)- 6 sin (π⁄3 ). sin (2π⁄3 )

f '(π⁄3 ) =3 (-1). (-0,5) - 6(½√3) ½√3

f '(π⁄3 ) =1,5 - 6 ( 1⁄4 3)

f '(π⁄3 ) = 1,5 - 4,5

f '(π⁄3 ) = -3

**Problem No.8**

Define the first instance of y = cos

^{3}(7x+3)

Answer:

y' = 3cos

y' = −21 cos

^{2}(7x+3) . -sin (7x+3) . 7y' = −21 cos

^{2}(7x+3) sin(7x+3)