The Trigonometric Functions by Stefan Waner and Steven R. Costenoble

This Section: 3. Derivatives of Trigonometric Functions

 2. The Six Trigonometric Functions Section 3 Exercises 4. Integrals of Trigonometric Functions Trigonometric Functions Main Page "Real World" Page

3. Derivatives of Trigonometric Functions

We shall start by giving the derivative of f(x) = sin x, and then using it to obtain the derivatives of the other five trigonometric functions.

Derivative of sin x

The derivative of the sine function is given by

 ddx sin x = cos x.

That's all there is to it!

Question
Where did that come from?

We shall justify this at the end of this section. (If you can't wait, press the pearl to go there now.)

Example 1

Calculate dy/dx if:

(a) y = x sin x (b) y = cosec x (c) y=
x2+x

sin x
(d) y = sin(3x21)

Solution

(a) An application of the calculator thought experiment (CTE) Ý tells us that x sin x is a product;

y = (x)(sin x).

Therefore, by the product rule,

 dydx = (1)(sin x) + (x)(cos x) = sin x + x cos x

(b) Recall from Section 2 that

 y = cosec x = 1sin x .

Therefore, by the quotient rule,

dy

dx
=  (0)(sin x) (1)(cos x) sin2x
(recall that sin2x is just (sin x)2)
=  cos xsin2x
=  cos xsin x . 1sin x
=  cotan x cosec x.
(from the identities in Section 2)

Notice that we have just obtained the derivative of one of the remaining five trigonometric functions. Four to go...

(c) Since the given function is a quotient,

 dydx = (2x+1)(sin x) (x2+x)(cos x) sin2x ,

and let us just leave it like that (there is no easy simplification of the answer).

(d) Here, an application of the CTEÝ tells us that y is the sine of a quantity.

Since

 ddx sin x = cos x,

the chain rule (press the pearl to go to the topic summary for a quick review) tells us that

 ddx sin u = cos u dudx

so that

d

dx
sin (3x21) =  cos (3x21) ddx (3x21)
= 6x cos(3x21) (we placed the 6x in front to avoid confusion-see below)

Ý See Example 6 on p. 258 in Calculus Applied to the Real World, or p. 756 in Finite Mathematics and Calculus Applied to the Real World. Alternatively, press here to consult the on-line topic summary, where the CTE is also discussed.

Before we go on...

Try to avoid writing expressions such as cos(3x21)(6x). Does this mean

 cos[(3x21)(6x)] (the cosine of the quantity (3x21)(6x)),

or does it mean

 [cos(3x21)](6x) (the product of cos(3x21) and 6x)?

That is why we placed the 6x in front of the cosine expression.

Question
What about the derivative of the cosine function?

Let us use the identity

cos x = sin(/2x)

from Section 1, and follow the method of Example 1(d) above: if

y = cos x = sin(/2x),

then, using the chain rule,

dy

dx
=  cos(/2x) ddx (/2x)
=  (1)cos(/2x)
(since /2 is constant, and d/dx(x) = 1)
= sin x (using the identity cos(/2x) = sin x)

Question
And the remaining three trigonometric functions?

Since all the remaining ones are expressible in terms of sin x and cos x, we shall leave them for you to do in the exercises!

Original Rule
Generalized Rule (Chain Rule)
 ddx sin x = cos x
 ddx sin u = cos u dudx
 ddx cos x = sin x
 ddx cos u = sin u dudx
 ddx tan x = sec2 x
 ddx tan u = sec2u dudx
 ddx cotan x = cosec2 x
 ddx cotan u = cosec2u dudx
 ddx sec x = sec x tan x
 ddx sec u = sec u tan u dudx
 ddx cosec x = cosec x cotan x
 ddx cosec u = cosec u cotan u dudx

Example 2

Find the derivatives of the following functions.

 (a) f(x) = tan(x21) (b) g(x) = cosec(e3x) (c) h(x) = exsin(2x) (d) r(x) = sin2x (e) s(x) = sin(x2)

Solution

(a) Since f(x) is the tan of a quantity, we use the chain rule form of the derivative of tangent:

 ddx tan u = sec2u dudx
 ddx tan(x21) = sec2(x21) d(x21)dx (substituting u = x21) = 2x sec2(x21).

(b) Since g(x) is the cosecant of a quantity, we use the rule

 ddx cosec u = cosec u cotan u dudx
 ddx cosec(e3x) = cosec(e3x) cotan(e3x) d(e3x)dx = 3e3x cosec(e3x) cotan(e3x). (the derivative of e3x is 3e3x)

(c) Since h(x) is the product of ex and sin(2x), we use the product rule,

h'(x) =  (ex)sin(2x) + ex ddx [sin(2x)]
=
(ex)sin(2x) + ex cos(2x)
d

dx
[2x]
(using d/dx sin u = cos u du/dx)
= exsin(2x) + 2excos(2x).

(d) Recall that sin2x = (sin x)2. Thus, r(x) is the square of a quantity (namely, the quantity sin x). Therefore, we use the chain rule for differentiating the square of a quantity,

 ddx [u2] = 2u dudx
 ddx [(sin x)2] = 2(sin x) d(sin x)dx = 2 sin x cos x.

(e) Notice the difference between sin2x and sin(x2). The first is the square of sin x, while the second is the sin of the quantity x2. Since we are differentiating the latter, we use the chain rule for differentiating the sine of a quantity:

 ddx sin u = cos u dudx
 ddx sin(x2) = cos(x2) d(x2)dx = 2x cos(x2).

Question
There is still some unfinished business...

Indeed. We will now motivate the formula that started it all:

 ddx sin x = cos x.

We shall do this calculation from scratch, using the general formula for a derivative:

 dd f(x) = h0 f(x+h) f(x)h

since here, f(x) = sin x, we can write

 dd sin(x) = h0 sin(x+h) sin(x)h . . . . (I)

We now use the addition formula in the preceding exercise set to expand sin(x+h):

sin(x+h) = sin x cos h + cos x sin h.

Substituting this in formula (I) gives

 dd sin(x) = h0 sin x cos h + cos x sin h sin(x)h .

Grouping the first and third terms together, and factoring out the sin x gives

d

d
sin(x) =  h0 sin x (cos h - 1) + cos x sin h h
=  h0 sin x (cos h - 1) h + h0 cos x sin hh
=  sin x h0 (cos h - 1) h + cos x h0 sin hh

and we are left with two limits to evaluate. Calculating these limits analytically requires a little trigonometry (press here for these calculations). Alternatively, we can get a good idea of what these two limits are by estimating them numerically. We find that:

 h0 (cos h 1) h = 0

and

 h0 sin h h = 1

Therefore,

 ddx sin x = sin x (0) + cos x (1) = cos x.

 2. The Six Trigonometric Functions Section 3 Exercises 4. Integrals of Trigonometric Functions Trigonometric Functions Main Page "Real World" Page

We would welcome comments and suggestions for improving this resource.

Mail us at:
 Stefan Waner (matszw@hofstra.edu) Steven R. Costenoble (matsrc@hofstra.edu)

Last Updated: March, 1997