Derivatives of the Trigonometric Functions

Derivative of f(x) = sin(x)

First note that angles will always be given in radians.  Degrees and calculus never go together.  If you ever hear the word "Degree" used in this class the appropriate question to ask is "Do you mean Celsius or Fahrenheit?"

We begin by exploring an important limit.  We can graph the function

sin x
f(x)  =
x

The graph is given below

The graph clearly indicates that as x approaches 0, y approaches 1.  We will use this fact in to find the derivative of f(x)  =  sin x.  The fact can be proven using triangles, but that should be done in a more advanced course.  Another important limit that help in the proof is

Once again, we will not prove this fact.  It can be found in more advanced textbooks.

 d                 sin x  =  cos x           dx

Proof:

We use the limit definition of the derivative along with the sum of angles formula for sin x.

Example

Find the derivative of

f(x)  =  x2 sin(3x)

Solution

We use the product and the chain rules.

f '(x)  =  (2x)(sin(3x)) + (x2)(3cos(3x))

=  2x sin(3x) + 3x2 cos(3x)

The Other Trigonometric Functions

We can find the derivatives of the other five trigonometric functions by using trig identities and rules of differentiation.  Below is a list of the six trig functions and their derivatives.

 f(x) f '(x) sin x cos x cos x -sin x tan x sec2 x sec x sec x tan x csc x -csc x cot x cot x -csc2 x

We will prove two of these.

The proof to the derivative of cos x

We use the identity

cos x  =  sin(p/2 - x)

Take the derivative of both sides using the chain rule for the derivative of the right hand side.

(cos x)'  =  - cos(p/2 - x)

Now use the identity

sin x  =  cos(p/2 - x)

to arrive at the result

(cos x)'  =  - sin x

The proof to the derivative of tan x

We use the quotient rule on

sin x
tan x  =
cos x

(cos x)(cos x) - (sin x)(-sin x)
(tan x)'  =
cos2 x

cos2 x + sin2 x                 1
=                                =                    =  sec2 x
cos2 x                      cos2 x

Application

The amount A of food available at day t of the year for a North American hare can be modeled by the equation

2p(t - 120)
A(t)  =  320 cos
365

On what day does the hare have the most food available?

Solution

We are asked to find a maximum, which involves taking the first derivative and setting it equal to zero.  We have

-320(2p)             2p(t - 120)
A'(t)  =                      sin
365                    365

Setting it equal to zero and solving gives us

2p(t - 120)
=  kp
365

When k  =  0, we get t  =  120 and when k  =  1  we get  t  =  302.5.  The second derivative test will tell us that t  =  120 gives us a maximum and t  =  320 gives us a minimum.  Other values of k will give us the same day of the year.  We can conclude that the hare has the greatest food supply on April 30.

Back to the Math Department Home