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En

  • one year ago

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

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  1. zepdrix
    • one year ago
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    Mmmmmmmmmmmmmmm

  2. zepdrix
    • one year ago
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    This one looks like a lot of work :\

  3. zepdrix
    • one year ago
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    You try it? :o Quotient rule, ya?

  4. En
    • one year ago
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    i did but my answer is not the same as from my book. It says that the answer should be zero.

  5. anonymous
    • one year ago
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    isn't 2x/x simply 2?

  6. zepdrix
    • one year ago
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    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

  7. anonymous
    • one year ago
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    oh is this all divided by

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

  9. zepdrix
    • one year ago
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    ?

  10. En
    • one year ago
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    what i mean to say is that the answer should be zero because the derivative of a constant is zero. i dont get it.

  11. anonymous
    • one year ago
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    is your question y=arctan(2x/x) or (arctan(2x))/x

  12. anonymous
    • one year ago
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    because it would make sense for the first one to have a derivative of zero

  13. hartnn
    • one year ago
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    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

  14. anonymous
    • one year ago
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    yeah

  15. anonymous
    • one year ago
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    fair call

  16. zepdrix
    • one year ago
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    So I'm confused... is the book saying the derivative should be zero? or you're saying that's what you think?

  17. anonymous
    • one year ago
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    thats unusual for that to have a derivative of zero

  18. En
    • one year ago
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    it really says the answer is zero.

  19. anonymous
    • one year ago
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    no other conditions?

  20. hartnn
    • one year ago
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    you don't get why derivative of a constant = 0 ??

  21. En
    • one year ago
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    yep. the only instruction given is : find the first derivative of the function given.

  22. anonymous
    • one year ago
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    aren't we assuming x is a variable though...

  23. zepdrix
    • one year ago
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    hmm, i dont like your book -_- maybe you're not telling us something lol

  24. anonymous
    • one year ago
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    lel

  25. En
    • one year ago
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    no, i get that the derivative of a constant is 0 , what i dont get is when should is recognize that its as constant.

  26. En
    • one year ago
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    * when should recognize it as a constant.

  27. anonymous
    • one year ago
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    perhaps there must be a condition for it to satisfy..

  28. En
    • one year ago
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    that the instructions

  29. zepdrix
    • one year ago
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    Awww man XD You didn't take a picture of this problem hehe!

  30. anonymous
    • one year ago
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    what am i even looking at..its so blury for me ahaha

  31. hartnn
    • one year ago
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    surely not 'differentiating w.r.t other variable' types of question... Answer given is 0 !?

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

  33. En
    • one year ago
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    no.19 guys....

  34. En
    • one year ago
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    *sigh.. guess i'll just pray this wont come out during exams tomorrow. </3

  35. anonymous
    • one year ago
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    where does the answer say its derivative is 0

  36. zepdrix
    • one year ago
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    ya hmm that's weird stuff goin on :D

  37. anonymous
    • one year ago
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    an exam just on trig functions?

  38. hartnn
    • one year ago
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    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

  39. SolomonZelman
    • one year ago
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    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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