I Find the constant value of the difference

[Mr Davis 97]

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...

 

[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.

 

[Mr Davis 97]

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.

 

[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.

 

[jostpuur]

The input x-1 is an unnecessary complication for the actual result, and we could also state that the formula

<br /> D_x \arctan\big(x + \sqrt{1 + x^2}\big) = \frac{1}{2}\frac{1}{1 + x^2}<br />

is true.