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Quick question on differentiation

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex] \frac{d}{dx} [ arctan (2x-3) ] [/tex]

    3. The attempt at a solution
    The general solution for d/dx arctan(x) = [tex] \frac{du/dx}{1+x^2} [/tex]

    so

    [tex] \frac{du/dx}{1+(2x-3)^2} [/tex]

    [tex] du/dx = 2 [/tex]

    [tex] \frac{2}{1+(2x-3)^2} [/tex]

    [tex] \frac{2}{1+2x^2-12x+9} [/tex]

    [tex] \frac{2}{10+2x^2-12x} [/tex]

    now this is where my question comes in,

    When I simplify it, I would divide all the terms in the numerator and denominator by 2,

    giving me [tex] \frac{1}{5+x^2-6x} [/tex]

    however i've been told that this simplification is wrong without any explanation

    why?
     
    Last edited: Aug 11, 2010
  2. jcsd
  3. Aug 11, 2010 #2

    Dick

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    It's wrong because (2x-3)^2=4*x^2-12x+9.
     
  4. Aug 11, 2010 #3
    Oh right, thanks

    also

    [tex] \frac{d}{dx} arcsin(2x-3) [/tex]

    =

    [tex] \frac{2}{\sqrt{1-(2x-3)^2}} [/tex]

    =

    [tex] \frac{2}{\sqrt{1-(4x^2-12x+9)}} [/tex]

    but apparently that's wrong too!
    ?

    it's supposed to be
    [tex] \frac{2}{\sqrt{1-(3-2x)^2}} [/tex]

    but i don't understand why,

    http://www.wolframalpha.com/input/?i=d/dx+arcsin(2x-3)
     
  5. Aug 11, 2010 #4

    Dick

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    Those two expressions are exactly the same thing. (2x-3)^2=(3-2x)^2=4x^2-12x+9. Are you overtired?
     
  6. Aug 11, 2010 #5
    just one of those days,
     
  7. Aug 11, 2010 #6
    Last one I promise,

    [tex]d/dx arcsin(\frac{2x+3}{5}) [/tex]

    I get from the equation, [tex] d/dx [ arcsin(x/a) ] = \frac{1}{\sqrt{a^2-x^2}} [/tex]

    [tex] \frac{2}{\sqrt{5^2-(2x+3)^2}} [/tex]

    =

    [tex] \frac{2}{\sqrt{25-9 - 4x^2 -12x}} [/tex]

    =

    [tex] \frac{2}{\sqrt{16-4x^2-12x}} [/tex]

    simplifying it by dividing everything by 2

    [tex] \frac{1}{\sqrt{8-2x^2-12x}} [/tex]

    however apparently i'm wrong again and could've simplified it further

    [tex] \frac{1}{\sqrt{x^2-3x+4}} [/tex]
    http://www.wolframalpha.com/input/?i=d/dx+arcsin((2x+3)/5)

    however I don't see how that's possible unless I have a 1/2 in the numerator

    is there a property of the square root that i'm missing?
     
  8. Aug 11, 2010 #7

    Dick

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    I really think you need a nap. Suppose I told you I'd simplified 2/sqrt(16) to 1/sqrt(8). Would you believe me? Are you sure you don't see what's wrong? You've got some other problems in there, but let's stick with that for now.
     
  9. Aug 11, 2010 #8
    I wouldn't believe you because 1/sqrt(8) != 2/4

    before I tried to simplify
    was the result

    [tex]
    \frac{2}{\sqrt{16-4x^2-12x}}
    [/tex]

    not correct?
    I thought I did that right. where would I go from here?
     
  10. Aug 11, 2010 #9

    Dick

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    That's correct. I would pull a factor of 4 out of the expression in the radical, then sqrt(4)=2 that cancels the 2 in the denominator. But this is getting silly. sqrt(a*b)=sqrt(a)*sqrt(b), right? Not sqrt(a*b)=a*sqrt(b). You MUST know that. You don't just pull factors out through powers.
     
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