2. The Six Trigonometric Functions  Section 3 Exercises  4. Integrals of Trigonometric Functions  Trigonometric Functions Main Page  "RealWorld" Page  Everything for Calculus  Español 
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) $y = \frac{x^2+x}{\sin x}$ 
(d) $y = \sin(3x^21)$ 
Solution
(a) An application of the calculator thought experiment (CTE)^{*} tells us that $x \sin x$ is a product;$y = (x)(\sin x).$
Therefore, by the product rule,
Therefore, by the quotient rule,
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,
and let us just leave it like that (there is no easy simplification of the answer).
(d) Here, an application of the CTE^{ } tells us that y is the sine of a quantity.
Since
the chain rule ( press the pearl to go to the topic summary for a quick review) tells us that
so that
^{*} 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 $\cos(3x^21)(6x).$ Does this mean
or does it mean
Question
What about the derivative of the cosine function?
Answer
Let us use the identity
$\cos x = \sin(\pi/2x)$
from Section 1, and follow the method of Example 1(d) above: if
$y = \cos x = \sin(\pi/2x),$
then, using the chain rule,
Question
And the remaining three trigonometric functions?
$\frac{d}{dx} \sin x = \cos x$  $\frac{d}{dx} \sin \color{blue}{u} = \cos \color{blue}{u}\frac{d\color{blue}{u}}{dx}$ 
$\frac{d}{dx} \cos x = \sin x$  $\frac{d}{dx} \cos \color{blue}{u} = \sin \color{blue}{u} \frac{d\color{blue}{u}}{dx}$ 
$\frac{d}{dx} \tan x = \sec^2 x$  $\frac{d}{dx} \tan \color{blue}{u} = \sec^2\color{blue}{u} \frac{d\color{blue}{u}}{dx}$ 
$\frac{d}{dx} \cotan x = \cosec^2 x$  $\frac{d}{dx} \cotan \color{blue}{u} = \cosec^2\color{blue}{u} \frac{d\color{blue}{u}}{dx}$ 
$\frac{d}{dx} \sec x = \sec x \tan x$  $\frac{d}{dx} \sec \color{blue}{u} = \sec \color{blue}{u}\ \tan \color{blue}{u} \frac{d\color{blue}{u}}{dx}$ 
$\frac{d}{dx} \cosec x = \cosec x \cotan x$  $\frac{d}{dx} \cosec \color{blue}{u} = \cosec \color{blue}{u}\ \cotan \color{blue}{u} \frac{d\color{blue}{u}}{dx}$ 
_{} Example 2
Find the derivatives of the following functions.
(a) $f(x) = \tan(x^21)$  (b) $g(x) = \cosec(e^{3x})$  (c) $h(x) = e^{x}\sin(2x)$ 
(d) $r(x) = \sin^2x$  (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:
$\frac{d}{dx} \tan(x^21) = \sec^2(x^21) \frac{d(x^21)}{dx}$  (substituting $u = x^21)$ 
$\ \ \ \ \ \ \ \ \ = 2x\ \sec^2(x^21).$ 
$\cosec(e^{3x})$  $= \cosec(e^{3x}) \cotan(e^{3x}) \frac{d(e^{3x})}{dx}$  
$= 3e^{3x} \cosec(e^{3x}) \cotan(e^{3x}).$  (the derivative of $e^{3x}$ is $3e^{3x})$ 
$h'(x) = (e^{x})\sin(2x) + e^{x} \frac{d}{dx} [\sin(2x)]$  
$\ \ \ \ \ \ \ \ \ = (e^{x}) \sin(2x) + e^{x} \cos(2x) \frac{d}{dx} [2x]$  (using $d/dx \sin u = \cos u\ du/dx)$ 
$\ \ \ \ \ \ \ \ \ = e^{x}\sin(2x) + 2e^{x}\cos(2x).$ 
$[(\sin x)^2]$  $=$  $2(\sin x)$  
$=$  $2 \sin x \cos x.$ 
$\sin(x^2)$  $=$  $\cos(x^2)$  
$=$  $2x \cos(x^2).$ 
Question
There is still some unfinished business...
Answer
Indeed. We will now motivate the formula that started it all:
We shall do this calculation from scratch, using the general formula for a derivative:
Substituting this in formula (I) gives
$\sin(x) = \lim_{h \to 0} \frac{\sin x (\cos h  1) + \cos x \sin h}{h}$  
$= \lim_{h \to 0} \frac{\sin x (\cos h  1)}{h} + \lim_{h \to 0} \frac{\cos x \sin h}{h}$  
$= \sin x\ \lim_{h \to 0} \frac{(\cos h  1)}{h} + \cos x\ \lim_{h \to 0} \frac{\sin h}{h}$ 
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:
and
Therefore,
2. The Six Trigonometric Functions  Section 3 Exercises  4. Integrals of Trigonometric Functions  Trigonometric Functions Main Page  "RealWorld" Page  Everything for Calculus 
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