Calculating dψ2/dx and d2ψ2/dx2

  • Thread starter dawozel
  • Start date
In summary, the conversation was about finding the second derivative of the function ψ2 = A(2αx2- 1)e-αx2/2 and how to properly use the product rule in the calculation. The correct second derivative was A(((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2)) + (-4ax)( \alpha x \exp(-\frac{1}{2}\alpha x^2)). It was also noted that it is important to double check the placement of parentheses when writing out the final equation.
  • #1
dawozel
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Homework Statement

ψ2 = A(2αx2- 1)e-αx2/2

First, calculate dψ2/dx, using A for A, x for x, and a for α.

Second, calculate d2ψ2/dx2.


3. The Attempt at a Solution
so I got the first derivative correct, it was

A((4*a*x)*exp((-a*x^2)/2) +(2*a*x^2 -1)*(-a*x*exp((-a*x^2)/2)))


but i can seem to calculate the second derivative correctly I'm getting
A((4*a)*(exp((-ax^2)/2)) + (4*a*x)*(-a*x*exp((-ax^2)/2)) +(4*a*x)*(-a*x*exp((-ax^2)/2)) +(2*a*x^2 -1)*(a^2*x^2*exp((-ax^2)/2)))

but this incorrect, am I missing something?
 
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  • #2
dawozel said:
Homework Statement

ψ2 = A(2αx2- 1)e-αx2/2
This is just about impossible to read. I could take a guess at what you're trying to say, but I shouldn't have to. Take a look at this, especially #2: https://www.physicsforums.com/showthread.php?t=617567.
dawozel said:
First, calculate dψ2/dx, using A for A, x for x, and a for α.

Second, calculate d2ψ2/dx2.


3. The Attempt at a Solution
so I got the first derivative correct, it was

A((4*a*x)*exp((-a*x^2)/2) +(2*a*x^2 -1)*(-a*x*exp((-a*x^2)/2)))


but i can seem to calculate the second derivative correctly I'm getting
A((4*a)*(exp((-ax^2)/2)) + (4*a*x)*(-a*x*exp((-ax^2)/2)) +(4*a*x)*(-a*x*exp((-ax^2)/2)) +(2*a*x^2 -1)*(a^2*x^2*exp((-ax^2)/2)))

but this incorrect, am I missing something?
 
  • #3
My bad this is the second derivative
[itex] A((4a)(exp(( - ax^2 ) / 2)) + (4ax) * ( - ax * exp(( - ax^2) /2)) + (4ax) * ( - a xexp(( - ax^2 ) / 2)) + (2ax - 1) (a^2x^2 *exp(( - ax^2 ) / 2))) [/itex]
 
  • #4
dawozel said:
My bad this is the second derivative
[itex] A((4a)(exp(( - ax^2 ) / 2)) + (4ax) * ( - ax * exp(( - ax^2) /2))[/itex]
[itex] + (4ax) * ( - a xexp(( - ax^2 ) / 2)) + (2ax - 1) (a^2x^2 *exp(( - ax^2 ) / 2))) [/itex]
That's a lot better.

Is this ψ2?

Here's what you have, cleaned up a little more, using more LaTeX and fewer parentheses.
$$A((4a)(e^{ - (a/2)x^2}) + (4ax) * ( - ax * e^{-(a/2)x^2}) + (4ax) * ( - a xe^{- (a/2)x^2}) + (2ax - 1) (a^2x^2 *e^{- (a/2)x^2}) $$

This ought to be at least close to what you have.
 
Last edited:
  • #5
Take a look at your last term, and consider the Product Rule. You missed something.
 
  • #6
This is the first derivative, correct?

##A[4axe^{-ax^2/2}+(2ax^2-1)(-axe^{-ax^2/2})]##

If so, in your computation of the second derivative, you need to perform a product rule within a product rule.
 
  • #7
yes that's my first derivative
 
  • #8
So my first derivative was simplified to [itex]\frac{d}{dx}\psi_2(x)=A\left[4\alpha x e^{-\frac{1}{2}\alpha x^2} - (2\alpha x^2 - 1)\alpha x e^{-\frac{1}{2}\alpha x^2}\right] [/itex]

and if i factor out the [itex] \alpha x \exp(-\frac{1}{2}\alpha x^2)[/itex]

i get that

[itex]\frac{d}{dx}\psi_2(x) = A(\alpha x \exp(-\frac{1}{2}\alpha x^2))(5-2ax^2) [/itex]

so my second derivative should be
[itex] A(((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2)) +(4ax)) [/itex]

is this right?
 
Last edited:
  • #9
dawozel said:
So my first derivative was simplified to [itex]\frac{d}{dx}\psi_2(x)=A\left[4\alpha x e^{-\frac{1}{2}\alpha x^2} - (2\alpha x^2 - 1)\alpha x e^{-\frac{1}{2}\alpha x^2}\right] [/itex]

and if i factor out the [itex] \alpha x \exp(-\frac{1}{2}\alpha x^2)[/itex]

i get that

[itex]\frac{d}{dx}\psi_2(x) = A(\alpha x \exp(-\frac{1}{2}\alpha x^2))(5-2ax^2) [/itex]

Looks good to me. Can you differentiate that now?
 
  • #10
so my second derivative should be
[itex] A(((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2)) +(4ax)) [/itex]

is this right or am i still missing a product rule?
 
  • #11
Still missing a bit. Try doing this in two steps. Let ##f(x) = \alpha x \exp(-\frac{1}{2}\alpha x^2)## and ##g(x) = 5-2ax^2##

What is the derivative of ##\frac{d}{dx}\psi_2(x) = Af(x)g(x)##? (simple application of product rule)

After that, calulate ##f'(x)## and ##g'(x)## and then plug everything into the the second derivative formula you got.

I know this is a lot of tedious work but I hope you'll be able to see why you were leaving out the terms you did.
 
  • #12
Hmmmm i may be forgetting to multiply the [itex] F' (x) by G(x) [/itex] and vice versa

Is the derivative closer to

[itex] A((((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2))(5-2ax^2) + (4ax)( \alpha x \exp(-\frac{1}{2}\alpha x^2))[/itex]
 
  • #13
That looks better. You were just forgetting to put in that very last term. One minor issue with a sign error, the derivative of ##5-2\alpha x^2## is ##-4\alpha x## so you need a minus sign in the one place.

And there's a bunch of parentheses; I'll just assume that they line up correctly. Just double check them if you are submitting them for homework.
 
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  • #14
Thanks for your help sir!
 

1. What is the mathematical definition of dψ2/dx and d2ψ2/dx2?

Both dψ2/dx and d2ψ2/dx2 are derivatives in calculus. dψ2/dx is the first derivative of the function ψ2 with respect to the variable x, while d2ψ2/dx2 is the second derivative of ψ2 with respect to x.

2. How do you calculate dψ2/dx and d2ψ2/dx2?

To calculate dψ2/dx, you need to use the limit definition of the derivative. This involves finding the limit as the change in x approaches 0. To calculate d2ψ2/dx2, you can use the same method, but instead of finding the limit of the first derivative, you find the limit of the second derivative.

3. What is the significance of dψ2/dx and d2ψ2/dx2 in calculus?

dψ2/dx and d2ψ2/dx2 are important in calculus because they represent the rate of change and the curvature of a function, respectively. They allow us to analyze the behavior of a function and make predictions about its future values.

4. When would you need to calculate dψ2/dx and d2ψ2/dx2 in a scientific experiment or study?

dψ2/dx and d2ψ2/dx2 may be needed in a scientific experiment or study when analyzing data that follows a certain pattern or trend. They can also be used to optimize a process or model a system.

5. Are there any applications of dψ2/dx and d2ψ2/dx2 outside of calculus?

Yes, dψ2/dx and d2ψ2/dx2 have applications in fields such as physics, engineering, and economics. They can be used to solve differential equations and model complex systems. They are also useful in optimization problems and in studying rates of change in real-world scenarios.

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