Find the constant value of the difference

  • #1
1,462
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I know that the following two functions have the same derivative: ##\arctan (x-1)## and ##2 \arctan (x-1 + \sqrt{(x-1)^2+1})##. Out of curiosity, how can I find the constant value at which they differ? I tried to add ##\pi / 2## to arctan(x-1) but I'm not sure if that works or not...
 

Answers and Replies

  • #2
member 587159
Don't know if this works, but if they are equal as functions, they are equal in every point on the domain. So try to find a good x to evaluate the functions in.
 
  • #3
1,462
44
Don't know if this works, but if they are equal as functions, they are equal in every point on the domain. So try to find a good x to evaluate the functions in.
So in that case does ##\arctan (x-1) + \pi / 2 =2 \arctan (x-1 + \sqrt{(x-1)^2+1})##? I'm not sure how to check it.
 
  • #4
member 587159
Let me be a little more specific. You know ##f'(x) = g'(x)##

Hence there exists a constant c such that ##f(x) = c + g(x)##

Now, you can start to find that constant by plugging in values of x.

In your example, put for example ##x = 1##.

Then ##0 = \arctan(0) = c + 2\arctan(1)##

Hence, ##c = -2 \arctan(1) = -2 \pi/4 = - \pi/2##

and indeed, ##\pi/2## works if you add it on the left.
 
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Likes Mr Davis 97
  • #5
2,111
18
The input [itex]x-1[/itex] is an unnecessary complication for the actual result, and we could also state that the formula

[tex]
D_x \arctan\big(x + \sqrt{1 + x^2}\big) = \frac{1}{2}\frac{1}{1 + x^2}
[/tex]

is true.
 
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