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Quick question on differentiation

  • Thread starter vorcil
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  • #1
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1. Homework Statement
[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:

Answers and Replies

  • #2
Dick
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It's wrong because (2x-3)^2=4*x^2-12x+9.
 
  • #3
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It's wrong because (2x-3)^2=4*x^2-12x+9.
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)
 
  • #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?
 
  • #5
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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?
just one of those days,
 
  • #6
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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?
 
  • #7
Dick
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Homework Helper
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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.
 
  • #8
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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.
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?
 
  • #9
Dick
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Homework Helper
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618
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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