Derivatives of Trigonometric Functions, Part 1

Post on 05-Dec-2014

277 views 0 download

description

In this presentation we calculate the derivative of sin(x). This is one of the basic formulas that is used to solve all the other trigonometric derivatives. For more lessons and videos: http://www.intuitive-calculus.com/derivatives-of-trigonometric-functions.html

Transcript of Derivatives of Trigonometric Functions, Part 1

Derivatives of Trigonometric Functions

1. f (x) = sin x

2. g(x) = cos x

Derivatives of Trigonometric Functions

The derivative of any trigonometric function can be found once weknow the derivative of the two basic functions:

1. f (x) = sin x

2. g(x) = cos x

Derivatives of Trigonometric Functions

The derivative of any trigonometric function can be found once weknow the derivative of the two basic functions:

1. f (x) = sin x

2. g(x) = cos x

Derivatives of Trigonometric Functions

The derivative of any trigonometric function can be found once weknow the derivative of the two basic functions:

1. f (x) = sin x

2. g(x) = cos x

Derivatives of Trigonometric Functions

The derivative of any trigonometric function can be found once weknow the derivative of the two basic functions:

1. f (x) = sin x

2. g(x) = cos x

In this video we’re going to find the derivative of sin x .

The derivative of sin x

The derivative of sin x

First of all, there is a little trig identity we need to remember:

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

Let’s now consider the function f (x) = sin x and let’s find itsderivative.

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

Let’s now consider the function f (x) = sin x and let’s find itsderivative.By definition:

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

Let’s now consider the function f (x) = sin x and let’s find itsderivative.By definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

Let’s now consider the function f (x) = sin x and let’s find itsderivative.By definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x= lim

∆x→0

sin(x + ∆x) − sin x

∆x

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

Let’s now consider the function f (x) = sin x and let’s find itsderivative.By definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x= lim

∆x→0

sin(x + ∆x) − sin x

∆x

Now we use the trig identity to expand sin(x + ∆x):

The derivative of sin x

First of all, there is a little trig identity we need to remember:

sin(a + b) = sin a cos b + sin b cos a

Let’s now consider the function f (x) = sin x and let’s find itsderivative.By definition:

f ′(x) = lim∆x→0

f (x + ∆x) − f (x)

∆x= lim

∆x→0

sin(x + ∆x) − sin x

∆x

Now we use the trig identity to expand sin(x + ∆x):

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

The derivative of sin x

The derivative of sin x

So, we need to solve this limit:

The derivative of sin x

So, we need to solve this limit:

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

The derivative of sin x

So, we need to solve this limit:

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

We can factor in the numerator:

The derivative of sin x

So, we need to solve this limit:

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

We can factor in the numerator:

f ′(x) = lim∆x→0

sin x (cos ∆x − 1) + sin ∆x cos x

∆x

The derivative of sin x

So, we need to solve this limit:

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

We can factor in the numerator:

f ′(x) = lim∆x→0

sin x (cos ∆x − 1) + sin ∆x cos x

∆x

= sin x lim∆x→0

cos ∆x − 1

∆x+ cos x lim

∆x→0

sin ∆x

∆x

The derivative of sin x

So, we need to solve this limit:

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

We can factor in the numerator:

f ′(x) = lim∆x→0

sin x (cos ∆x − 1) + sin ∆x cos x

∆x

= sin x lim∆x→0

cos ∆x − 1

∆x+ cos x

�������*

1

lim∆x→0

sin ∆x

∆x

The derivative of sin x

So, we need to solve this limit:

f ′(x) = lim∆x→0

sin x cos ∆x + sin ∆x cos x − sin x

∆x

We can factor in the numerator:

f ′(x) = lim∆x→0

sin x (cos ∆x − 1) + sin ∆x cos x

∆x

= sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x��

�����*

1

lim∆x→0

sin ∆x

∆x

The derivative of sin x

The derivative of sin x

So, we have that:

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x=

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x=

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

1 − cos2 ∆x

∆x (1 + cos ∆x)

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= − lim∆x→0

sin ∆x

∆x. lim

∆x→0

sin ∆x

1 + cos ∆x

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= −���

����*1

lim∆x→0

sin ∆x

∆x. lim

∆x→0

sin ∆x

1 + cos ∆x

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= −���

����*1

lim∆x→0

sin ∆x

∆x. lim

∆x→0

����: 0

sin ∆x

1 + cos ∆x

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= −���

����*1

lim∆x→0

sin ∆x

∆x. lim

∆x→0

����: 0

sin ∆x

1 +����: 1

cos ∆x

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= −���

����*1

lim∆x→0

sin ∆x

∆x.���

������:

0lim

∆x→0

sin ∆x

1 + cos ∆x

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= −���

����*1

lim∆x→0

sin ∆x

∆x.���

������:

0lim

∆x→0

sin ∆x

1 + cos ∆x= 0

The derivative of sin x

So, we have that:

f ′(x) = sin x lim∆x→0

cos ∆x − 1

∆x︸ ︷︷ ︸We’ll show this is 0!

+ cos x

lim∆x→0

cos ∆x − 1

∆x= − lim

∆x→0

1 − cos ∆x

∆x=

= − lim∆x→0

1 − cos ∆x

∆x.1 + cos ∆x

1 + cos ∆x= − lim

∆x→0

�����

��: sin2 ∆x

1 − cos2 ∆x

∆x (1 + cos ∆x)

= −�������*1

lim∆x→0

sin ∆x

∆x.���

������:

0lim

∆x→0

sin ∆x

1 + cos ∆x= 0

The derivative of sin x

The derivative of sin x

So, finally:

The derivative of sin x

So, finally:

f ′(x) = cos x

The derivative of sin x

So, finally:

f ′(x) = cos x

Another way to put it:

The derivative of sin x

So, finally:

f ′(x) = cos x

Another way to put it:

d

dx(sin x) = cos x

The derivative of sin x

So, finally:

f ′(x) = cos x

Another way to put it:

d

dx(sin x) = cos x