Solving Second Diff. Homework: y"(x)

[Jenkz]

Homework Statement

y(x) = exp (-(\sqrt{ms}/2t) x^{2})

Find the y"(x)

The Attempt at a Solution

y'(x) = (-(\sqrt{ms}/2t) 2x) exp (-(\sqrt{ms}/2t) x^{2})

y"(x) = (-(\sqrt{ms}/t)) exp (-(\sqrt{ms}/2t) x^{2}) + ((ms/4t^{2})4x^{2}) exp (-(\sqrt{ms}/2t) x^{2})This is correct? Sorry if it looks a bit messy... Thanks.

 

[theJorge551]

Could you please try to clean up the equation of the problem a little bit? Just for clarification.

 

[Jenkz]

I hope that makes it easier.

 

[epenguin]

Could you please try to clean up the equation of the problem a little bit? Just for clarification.

I hope that makes it easier.

I can't see any that yet.

I think Jorge's point is, contrary to what many students imagine, there is no virtue in complicated-looking expressions, in carrying through all unnecessary complication in a problem until, maybe at the end, it matters. You have, if y'(x) means dy/dx, a function of ( x² multiplied by a constant). You don't need to know how the constant is made up of this, that and the other when you differentiate. So just call it a. Or you can call it -a. Then you can see what you are doing easier and make fewer mistakes.

As you go through phys and math you will see all the time where where where. I.e. w = some function of, say, [au + sin²(bv)] where a and b each = some other jumble of constants stuff (sometimes quite complicated stuff, like 'where a is the solution of this horrible equation' - something you could never carry through with everything explicit all the time). At the end of a calculation you might need to unravel or put back what is in the a and b to see how, e.g. a physical behaviour depends on the parameters inside them.

 

[Mark44]

I think Jorge's point is, contrary to what many students imagine, there is no virtue in complicated-looking expressions, in carrying through all unnecessary complication in a problem until, maybe at the end, it matters. You have, if y'(x) means dy/dx, a function of ( x² multiplied by a constant). You don't need to know how the constant is made up of this, that and the other when you differentiate. So just call it a. Or you can call it -a. Then you can see what you are doing easier and make fewer mistakes.

Excellent point. After all, \sqrt{ms}/2t is just a constant as far as differentiation with respect to x is concerned.

BTW, you (the OP) are trying to find the second derivative, not the second differential. Also, this is hardly a precalculus problem.

 

[Jenkz]

Thank you for the advice.

@Mark44: sorry about that.