En
  • En
PLEASE HELP! Find the fist derivative of y=Arctan 2x/x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
Mmmmmmmmmmmmmmm
zepdrix
  • zepdrix
This one looks like a lot of work :\
zepdrix
  • zepdrix
You try it? :o Quotient rule, ya?

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En
  • En
i did but my answer is not the same as from my book. It says that the answer should be zero.
anonymous
  • anonymous
isn't 2x/x simply 2?
zepdrix
  • zepdrix
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
anonymous
  • anonymous
oh is this all divided by
En
  • En
the same thing goes for the problem...\[w \frac{ \arcsin a }{ a }\] it says that the derivative of a constant is zero...
zepdrix
  • zepdrix
?
En
  • En
what i mean to say is that the answer should be zero because the derivative of a constant is zero. i dont get it.
anonymous
  • anonymous
is your question y=arctan(2x/x) or (arctan(2x))/x
anonymous
  • anonymous
because it would make sense for the first one to have a derivative of zero
hartnn
  • hartnn
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
anonymous
  • anonymous
yeah
anonymous
  • anonymous
fair call
zepdrix
  • zepdrix
So I'm confused... is the book saying the derivative should be zero? or you're saying that's what you think?
anonymous
  • anonymous
thats unusual for that to have a derivative of zero
En
  • En
it really says the answer is zero.
anonymous
  • anonymous
no other conditions?
hartnn
  • hartnn
you don't get why derivative of a constant = 0 ??
En
  • En
yep. the only instruction given is : find the first derivative of the function given.
anonymous
  • anonymous
aren't we assuming x is a variable though...
zepdrix
  • zepdrix
hmm, i dont like your book -_- maybe you're not telling us something lol
anonymous
  • anonymous
lel
En
  • En
no, i get that the derivative of a constant is 0 , what i dont get is when should is recognize that its as constant.
En
  • En
* when should recognize it as a constant.
anonymous
  • anonymous
perhaps there must be a condition for it to satisfy..
En
  • En
that the instructions
zepdrix
  • zepdrix
Awww man XD You didn't take a picture of this problem hehe!
anonymous
  • anonymous
what am i even looking at..its so blury for me ahaha
hartnn
  • hartnn
surely not 'differentiating w.r.t other variable' types of question... Answer given is 0 !?
fouzberzerk
  • fouzberzerk
Can you guys help me on my question? I've been having some issues and no one is responding.
En
  • En
no.19 guys....
En
  • En
*sigh.. guess i'll just pray this wont come out during exams tomorrow.
anonymous
  • anonymous
where does the answer say its derivative is 0
zepdrix
  • zepdrix
ya hmm that's weird stuff goin on :D
anonymous
  • anonymous
an exam just on trig functions?
hartnn
  • hartnn
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
SolomonZelman
  • SolomonZelman
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!

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