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

That's all there is to it!
Question
Where did that come from?
Answer
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  =  sin x 
(d) y = sin(3x^{2}1) 
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,
dx  =  (1)(sin x) + (x)(cos x) = sin x + x cos x 
(b) Recall from Section 2 that
y = cosec x =  sin x  . 
Therefore, by the quotient rule,
dx 
= 

(recall that sin^{2}x is just (sin x)^{2})  
= 


= 


= 

(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,
dx  =  sin^{2}x  , 
and let us just leave it like that (there is no easy simplification of the answer).
(d) Here, an application of the CTE^{Ý} tells us that y is the sine of a quantity.
Since
dx  sin x = cos x, 
the chain rule (press the pearl to go to the topic summary for a quick review) tells us that
dx  sin u = cos u  dx 
so that
dx 
sin (3x^{2}1)  = 


=  6x cos(3x^{2}1)  (we placed the 6x in front to avoid confusionsee 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 online topic summary, where the CTE is also discussed.
Before we go on...
Try to avoid writing expressions such as cos(3x^{2}1)(6x). Does this mean
cos[(3x^{2}1)(6x)]  (the cosine of the quantity (3x^{2}1)(6x)), 
or does it mean
[cos(3x^{2}1)](6x)  (the product of cos(3x^{2}1) and 6x)? 
That is why we placed the 6x in front of the cosine expression.
Question
What about the derivative of the cosine function?
Answer
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,
dx 
= 


= 

(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?
Answer
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!

















_{} Example 2
Find the derivatives of the following functions.
(a) f(x) = tan(x^{2}1)  (b) g(x) = cosec(e^{3x})  (c) h(x) = e^{x}sin(2x) 
(d) r(x) = sin^{2}x  (e) s(x) = sin(x^{2}) 
Solution
(a) Since f(x) is the tan of a quantity, we use the chain rule form of the derivative of tangent:
dx 
tan u = sec^{2}u  dx 
dx 
tan(x^{2}1)  =  sec^{2}(x^{2}1)  dx 
(substituting u = x^{2}1) 
=  2x sec^{2}(x^{2}1). 
(b) Since g(x) is the cosecant of a quantity, we use the rule
dx 
cosec u = cosec u cotan u  dx 
dx 
cosec(e^{3x})  =  cosec(e^{3x}) cotan(e^{3x})  dx 

=  3e^{3x} cosec(e^{3x}) cotan(e^{3x}).  (the derivative of e^{3x} is 3e^{3x}) 
(c) Since h(x) is the product of e^{x} and sin(2x), we use the product rule,
h'(x)  = 


= 

(using d/dx sin u = cos u du/dx)  
=  e^{x}sin(2x) + 2e^{x}cos(2x). 
(d) Recall that sin^{2}x = (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,
dx 
[u^{2}]  =  2u  dx 
dx 
[(sin x)^{2}]  =  2(sin x)  dx 
=  2 sin x cos x. 
(e) Notice the difference between sin^{2}x and sin(x^{2}). The first is the square of sin x, while the second is the sin of the quantity x^{2}. Since we are differentiating the latter, we use the chain rule for differentiating the sine of a quantity:
dx 
sin u = cos u  dx 
dx 
sin(x^{2})  =  cos(x^{2})  dx 
=  2x cos(x^{2}). 
Question
There is still some unfinished business...
Answer
Indeed. We will now motivate the formula that started it all:
dx 
sin x  =  cos x. 
We shall do this calculation from scratch, using the general formula for a derivative:
d 
f(x) =  h0  h 
since here, f(x) = sin x, we can write
d 
sin(x) =  h0  h 
. . . . (I) 
We now use the addition formula in the preceding exercise set to expand sin(x+h):
Substituting this in formula (I) gives
d 
sin(x) =  h0  h 
. 
Grouping the first and third terms together, and factoring out the sin x gives
d 
sin(x)  = 


= 


= 

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  h 
=  0 
and
h0  h 
=  1 
Therefore,
dx 
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 
Mail us at:
Stefan Waner (matszw@hofstra.edu)  Steven R. Costenoble (matsrc@hofstra.edu) 