Question

PLEASE HELP!
Find the fist derivative of y=Arctan 2x/x

Answer 1

Mmmmmmmmmmmmmmm

Answer 2

This one looks like a lot of work :\

Answer 3

You try it? :o
Quotient rule, ya?

Answer 4

i did but my answer is not the same as from my book. It says that the answer should be zero.

Answer 5

isn't 2x/x simply 2?

Answer 6

Quotient rule :P\[\large\rm \left(\frac{\arctan2x}{x}\right)'=\frac{(\arctan2x)'x-\arctan2x(x)'}{(x)^2}\]And then derivative,\[\large\rm =\frac{\frac{2}{1+(2x)^2}-\arctan2x(1)}{x^2}\]And then it's just a bunch of simplification from there, ya? :o

Answer 7

oh is this all divided by

Answer 8

the same thing goes for the problem...\[w \frac{ \arcsin a }{ a }\] it says that the derivative of a constant is zero...

Answer 9

?

Answer 10

what i mean to say is that the answer should be zero because the derivative of a constant is zero.


i dont get it.

Answer 11

is your question y=arctan(2x/x) or (arctan(2x))/x

Answer 12

because it would make sense for the first one to have a derivative of zero

Answer 13

then the question must not be to find the derivative with respect to x.

because if the variable is not x
then y is constant

Answer 14

yeah

Answer 15

fair call

Answer 16

So I'm confused...
is the book saying the derivative should be zero?
or you're saying that's what you think?

Answer 17

thats unusual for that to have a derivative of zero

Answer 18

it really says the answer is zero.

Answer 19

no other conditions?

Answer 20

you don't get why derivative of a constant = 0 ??

Answer 21

yep. the only instruction given is : find the first derivative of the function given.

Answer 22

aren't we assuming x is a variable though...

Answer 23

hmm, i dont like your book -_-
maybe you're not telling us something lol

Answer 24

lel

Answer 25

no, i get that the derivative of a constant is 0 , what i dont get is when should is recognize that its as constant.

Answer 26

* when should recognize it as a constant.

Answer 27

perhaps there must be a condition for it to satisfy..

Answer 28

that the instructions

Answer 29

Awww man XD
You didn't take a picture of this problem hehe!

Answer 30

what am i even looking at..its so blury for me ahaha

Answer 31

surely not 'differentiating w.r.t other variable'
types of question...

Answer given is 0 !?

Answer 32

Can you guys help me on my question? I've been having some issues and no one is responding.

Answer 33

no.19 guys....

Answer 34

*sigh.. guess i'll just pray this wont come out during exams tomorrow. </3

Answer 35

where does the answer say its derivative is 0

Answer 36

ya hmm that's weird stuff goin on :D

Answer 37

an exam just on trig functions?

Answer 38

i'd say,
just so you get some practice,
ignore that the answer is 0.

start solving that problem using quotient rule,
which I am quite sure you know

Answer 39

Perhaps a similar example would help.
I will assume that you know other properties
of exponents, trigonometric functions, and etc..

So, lets say you have the following function (below), which
you are assigned to differentiate (with respect to x).

\(\large\color{blue}{ \displaystyle f(x)=\frac{{\rm Arctan}{(3x)}}{3x} }\)

To find the derivative, which you probably know to denote by \(f'(x)\),
we would first find the derivative of \({\rm Arctan}{(3x)}\), and the derivative
of \(3x\) you already know. Then, we can easily use the quotient rule.
---------------------------------------------------

\(\large\color{blue}{ \displaystyle y={\rm Arctan}{(3x)} }\)

\(\large\color{blue}{ \displaystyle 3x={\rm tan}{(y)} }\)

(Note, that we are differentiating with respect to x, where y
is a function of x. For this reason any derivative of a "y" is
going to have a chain rule, and that is: \(y'\).)

\(\large\color{blue}{ \displaystyle 3={\rm sec}^2{(y)}\times y' }\)

The derivative of 3x is 3. The derivative of \(\tan(x)\) is \(\sec^2(x)\), but here,
since we have the argument of \(y\), we have to multiply times \(y'\), as I noted
previously.

We are dividing by \(\sec^2(x)\) on both sides.
(You will see why I didn't just simply say \(3\cos^2(x)\)
on the right side)

\(\large\color{blue}{ \displaystyle y'=\frac{3}{\sec^2(x)}\ }\)

Recall that: \(\sec^2\theta=\tan^2\theta+1\), and when we apply this, we get:

\(\large\color{blue}{ \displaystyle y'=\frac{3}{\tan^2(x)+1}\ }\)

Now, you have to go back to the very beginning, because there we
had that: \(\color{black}{ \displaystyle {\rm tan}{(y)} =3x}\). And therefore, when you square both sides,
it comes out that: \( \tan^2(y)=9x^2\).

So, we got to substitute that, and we have our final answer:

\(\large\color{blue}{ \displaystyle y'=\frac{3}{9x^2+1}\ }\)

This is the derivative of \({\rm Arctan}(x)\).
---------------------------------------------------

Okay, now back to our initial function:

\(\large\color{red}{ \displaystyle f(x)=\frac{{\rm Arctan}{(3x)}}{3x} }\)

Note, that the quotient rule is: \({\\[0.5em]}\)
\(\color{black}{\displaystyle \frac{dy}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{\left[~g(x)~\right]^2}}\)
And when we apply this, we get:

\(\large\color{red}{ \displaystyle f'(x)=\frac{3x \cdot \left(\frac{3}{9x^2+1}\right)-3{\rm Arctan}{(3x)} }{(3x)^2} }\)

\(\large\color{red}{ \displaystyle f'(x)=\frac{\frac{9x^2}{9x^2+1}-3{\rm Arctan}{(3x)} }{9x^2} }\)

\(\large\color{red}{ \displaystyle f'(x)=\frac{\frac{9x^2}{9x^2+1} }{9x^2} -\frac{3{\rm Arctan}{(3x)} }{9x^2} }\)

\(\large\color{red}{ \displaystyle f'(x)=\frac{\frac{1}{9x^2+1} }{1} -\frac{{\rm Arctan}{(3x)} }{3x^2} }\)

\(\large{\bbox[5pt, lightyellow ,border:2px solid black ]{ \displaystyle \color{red}{f'(x)=\frac{1}{9x^2+1}-\frac{{\rm Arctan}{(3x)} }{3x^2}} }}\)

I have just performed some minor algebraic tasks to simplify my derivative,
but I hope this example is helpful somewhat.

Good Luck!