# 3.5 Derivatives of trigonometric functions  (Page 2/3)

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## Differentiating a function containing sin x

Find the derivative of $f\left(x\right)=5{x}^{3}\text{sin}\phantom{\rule{0.1em}{0ex}}x.$

Using the product rule, we have

$\begin{array}{cc}\hfill f\prime \left(x\right)& =\frac{d}{dx}\left(5{x}^{3}\right)·\text{sin}\phantom{\rule{0.1em}{0ex}}x+\frac{d}{dx}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)·5{x}^{3}\hfill \\ & =15{x}^{2}·\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x·5{x}^{3}.\hfill \end{array}$

After simplifying, we obtain

${f}^{\prime }\left(x\right)=15{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x+5{x}^{3}\text{cos}\phantom{\rule{0.1em}{0ex}}x.$

Find the derivative of $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x.$

${f}^{\prime }\left(x\right)={\text{cos}}^{2}x-{\text{sin}}^{2}x$

## Finding the derivative of a function containing cos x

Find the derivative of $g\left(x\right)=\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{4{x}^{2}}.$

By applying the quotient rule, we have

${g}^{\prime }\left(x\right)=\frac{\left(\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)4{x}^{2}-8x\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}{{\left(4{x}^{2}\right)}^{2}}.$

Simplifying, we obtain

$\begin{array}{cc}\hfill {g}^{\prime }\left(x\right)& =\frac{-4{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x-8x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x}{16{x}^{4}}\hfill \\ & =\frac{\text{−}x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x-2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x}{4{x}^{3}}.\hfill \end{array}$

Find the derivative of $f\left(x\right)=\frac{x}{\text{cos}\phantom{\rule{0.1em}{0ex}}x}.$

$\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x+x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{{\text{cos}}^{2}x}$

## An application to velocity

A particle moves along a coordinate axis in such a way that its position at time $t$ is given by $s\left(t\right)=2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t-t$ for $0\le t\le 2\pi .$ At what times is the particle at rest?

To determine when the particle is at rest, set ${s}^{\prime }\left(t\right)=v\left(t\right)=0.$ Begin by finding ${s}^{\prime }\left(t\right).$ We obtain

${s}^{\prime }\left(t\right)=2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t-1,$

so we must solve

$2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t-1=0\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}0\le t\le 2\pi .$

The solutions to this equation are $t=\frac{\pi }{3}$ and $t=\frac{5\pi }{3}.$ Thus the particle is at rest at times $t=\frac{\pi }{3}$ and $t=\frac{5\pi }{3}.$

A particle moves along a coordinate axis. Its position at time $t$ is given by $s\left(t\right)=\sqrt{3}t+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t$ for $0\le t\le 2\pi .$ At what times is the particle at rest?

$t=\frac{\pi }{3},t=\frac{2\pi }{3}$

## Derivatives of other trigonometric functions

Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives.

## The derivative of the tangent function

Find the derivative of $f\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}x.$

Start by expressing $\text{tan}\phantom{\rule{0.1em}{0ex}}x$ as the quotient of $\text{sin}\phantom{\rule{0.1em}{0ex}}x$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x:$

$f\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}x=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{\text{cos}\phantom{\rule{0.1em}{0ex}}x}.$

Now apply the quotient rule to obtain

${f}^{\prime }\left(x\right)=\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x-\left(\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)\text{sin}\phantom{\rule{0.1em}{0ex}}x}{{\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}^{2}}.$

Simplifying, we obtain

${f}^{\prime }\left(x\right)=\frac{{\text{cos}}^{2}x+{\phantom{\rule{0.2em}{0ex}}\text{sin}}^{2}x}{{\text{cos}}^{2}x}.$

Recognizing that ${\text{cos}}^{2}x+{\text{sin}}^{2}x=1,$ by the Pythagorean theorem, we now have

${f}^{\prime }\left(x\right)=\frac{1}{{\text{cos}}^{2}x}.$

Finally, use the identity $\text{sec}\phantom{\rule{0.1em}{0ex}}x=\frac{1}{\text{cos}\phantom{\rule{0.1em}{0ex}}x}$ to obtain

${f}^{\prime }\left(x\right)={\text{sec}}^{2}x.$

Find the derivative of $f\left(x\right)=\text{cot}\phantom{\rule{0.1em}{0ex}}x.$

${f}^{\prime }\left(x\right)=\text{−}{\text{csc}}^{2}x$

The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem.

## Derivatives of $\text{tan}\phantom{\rule{0.1em}{0ex}}x,\text{cot}\phantom{\rule{0.1em}{0ex}}x,\text{sec}\phantom{\rule{0.1em}{0ex}}x,$ And $\text{csc}\phantom{\rule{0.1em}{0ex}}x$

The derivatives of the remaining trigonometric functions are as follows:

$\frac{d}{dx}\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)={\text{sec}}^{2}x$
$\phantom{\rule{0.72em}{0ex}}\frac{d}{dx}\left(\text{cot}\phantom{\rule{0.1em}{0ex}}x\right)=\text{−}{\text{csc}}^{2}x$
$\phantom{\rule{1.4em}{0ex}}\frac{d}{dx}\left(\text{sec}\phantom{\rule{0.1em}{0ex}}x\right)=\text{sec}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x$
$\phantom{\rule{2.2em}{0ex}}\frac{d}{dx}\left(\text{csc}\phantom{\rule{0.1em}{0ex}}x\right)=\text{−}\text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}\mathrm{x.}$

## Finding the equation of a tangent line

Find the equation of a line tangent to the graph of $f\left(x\right)=\text{cot}\phantom{\rule{0.1em}{0ex}}x$ at $x=\frac{\text{π}}{4}.$

To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute

$f\left(\frac{\pi }{4}\right)=\text{cot}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{4}=1.$

Thus the tangent line passes through the point $\left(\frac{\pi }{4},1\right).$ Next, find the slope by finding the derivative of $f\left(x\right)=\text{cot}\phantom{\rule{0.1em}{0ex}}x$ and evaluating it at $\frac{\pi }{4}\text{:}$

${f}^{\prime }\left(x\right)=\text{−}{\text{csc}}^{2}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(\frac{\pi }{4}\right)=\text{−}{\text{csc}}^{2}\left(\frac{\pi }{4}\right)=-2.$

Using the point-slope equation of the line, we obtain

$y-1=-2\left(x-\frac{\pi }{4}\right)$

or equivalently,

$y=-2x+1+\frac{\pi }{2}.$

## Finding the derivative of trigonometric functions

Find the derivative of $f\left(x\right)=\text{csc}\phantom{\rule{0.1em}{0ex}}x+x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x.$

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

${f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\text{csc}\phantom{\rule{0.1em}{0ex}}x\right)+\frac{d}{dx}\left(x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x\right).$

In the first term, $\frac{d}{dx}\left(\text{csc}\phantom{\rule{0.1em}{0ex}}x\right)=\text{−}\text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x,$ and by applying the product rule to the second term we obtain

$\frac{d}{dx}\left(x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)=\left(1\right)\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)+\left({\text{sec}}^{2}x\right)\left(x\right).$

Therefore, we have

${f}^{\prime }\left(x\right)=\text{−}\text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x+\text{tan}\phantom{\rule{0.1em}{0ex}}x+x\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}x.$

why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal