# Thread: Manipulation of the constant C that ocurres after differentiating

1. ## Manipulation of the constant C that ocurres after differentiating

Hi!

I have a question about something I find a bit peculiar. The textbook of my math-course is a bit inconistent in the use of the constant (here $\displaystyle C$) that comes from differentiating an expression.

I have an expression that has been differentiated, and it looks like this:

$\displaystyle y = \sqrt{e^{\frac{1}{2}x^2+C}}$

Sometimes this kind of expression is used to calculate further, but other times it is manipulated into this:

$\displaystyle y = \sqrt{Ce^{\frac{1}{2}x^2}}$

My question is this:
Can I always perfrom this manipulation, or is it just in certain circumstances? I know the constant $\displaystyle C$ is a constant that can be changed if it is only affected by other constants (like $\displaystyle e$ in this situation), but the whole thing is very confusing...

Any clarifying help will be greatly appreciated!

Thomas

2. Originally Posted by luckytommy
Hi!

I have a question about something I find a bit peculiar. The textbook of my math-course is a bit inconistent in the use of the constant (here $\displaystyle C$) that comes from differentiating an expression.

I have an expression that has been differentiated, and it looks like this:

$\displaystyle y = \sqrt{e^{\frac{1}{2}x^2+C}}$

Sometimes this kind of expression is used to calculate further, but other times it is manipulated into this:

$\displaystyle y = \sqrt{Ce^{\frac{1}{2}x^2}}$

My question is this:
Can I always perfrom this manipulation, or is it just in certain circumstances? I know the constant $\displaystyle C$ is a constant that can be changed if it is only affected by other constants (like $\displaystyle e$ in this situation), but the whole thing is very confusing...

Any clarifying help will be greatly appreciated!

Thomas
$\displaystyle e^{f(x) + C} = e^{f(x)} e^C = A e^{f(x)}$.

3. Originally Posted by mr fantastic
$\displaystyle e^{f(x) + C} = e^{f(x)} e^C = A e^{f(x)}$.
Yes, I just skipped those steps in my post. The thing is that the book we use sometimes gives;
$\displaystyle e^{\frac{1}{2}x^2+C}$
as an eligible solution, while other times;
$\displaystyle Ce^{\frac{1}{2}x^2}$
is preferred. I still don't know why the authors would do this. Maybe it is caused by the simple fact that I am writing authors instead of author ??

Anyway: I have understood that this "maneuver" is always allowed (at least from problems I have seen ), and that I prefer the latter form.

4. As explained in...

http://www.mathhelpforum.com/math-he...tegration.html

... the problem is when in the expression...

$\displaystyle c\cdot e^{\frac{x^{2}}{2}}$ (1)

... is $\displaystyle c=0$ [or also $\displaystyle c\le 0$...]. Clearly in this case doesn't hold the identity...

$\displaystyle c\cdot e^{\frac{x^{2}}{2}}= e^{\frac{x^{2}}{2}+\ln c}$ (2)

... because $\displaystyle \ln c$ doesn't exist ...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

5. Okay, I understand. So in these situations the "maneuver" would give the one form a smaller specter of solutions than the other...