1. ## Exponential Differentiation Help!

Hello,

I'm currently doing an bridging course and the people who I usually ask for help have time off for Christmas. As such I am hoping to get some assistance here.

I am having trouble working out the second derivative of the following function: y=Ate^(-bt) with t being a variable, and A and B being unknown constants. The question I am solving is, "Verify the point of inflection occurs at t=2/b."

I have already found the first derivative to be A*(e^(-bt))-(bte^(-bt)), and assuming that is correct, I have worked out the following so far for the second derivative:

=A*(d/dt(e^(-bt)))-(b*(d/dt(t)*e^(-bt)))
=A*(e^(-bt)*(d/dt(-bt)))-(b*(d/dt(t)*e^(-bt))+(t*(d/dt(e^(-bt))))
=A*(-be^(-bt))-be^(-bt)+(t*e^(-bt)*(d/dt(-bt)))

After this point I am getting myself confused. I have been using http://www.derivative-calculator.net...=2&showsteps=1 as a guide, but am having trouble figuring out why the (d/dt(t)) becomes a +1, instead of just being a multiplier in the third to last line.

Once I get this derivative worked out, I then need to solve to 0 to find the PoI.

Any help will be appreciated.

Thanks.

2. Originally Posted by Whitey
Hello,

I'm currently doing an bridging course and the people who I usually ask for help have time off for Christmas. As such I am hoping to get some assistance here.

I am having trouble working out the second derivative of the following function: y=Ate^(-bt) with t being a variable, and A and B being unknown constants. The question I am solving is, "Verify the point of inflection occurs at t=2/b."

I have already found the first derivative to be A*(e^(-bt))-(bte^(-bt)), and assuming that is correct, I have worked out the following so far for the second derivative:

=A*(d/dt(e^(-bt)))-(b*(d/dt(t)*e^(-bt)))
=A*(e^(-bt)*(d/dt(-bt)))-(b*(d/dt(t)*e^(-bt))+(t*(d/dt(e^(-bt))))
=A*(-be^(-bt))-be^(-bt)+(t*e^(-bt)*(d/dt(-bt)))

After this point I am getting myself confused. I have been using http://www.derivative-calculator.net...=2&showsteps=1 as a guide, but am having trouble figuring out why the (d/dt(t)) becomes a +1, instead of just being a multiplier in the third to last line.

Once I get this derivative worked out, I then need to solve to 0 to find the PoI.

Any help will be appreciated.

Thanks.
That is incorrect.

d/dt[A*t*e(-bt)] = A * d/dt[t * e(-bt)] = A * [t * (-b) * e(-bt) + e(-bt)] = A * e(-bt) * [1 - bt]

d2/dt2[A*t*e(-bt)] = d/dt {A * [1 - bt]* e(-bt)} = A * d/dt {[1 - bt]* e(-bt)} = A * {[1-bt]*(-b) * e(-bt) + (-b) * e(-bt)} = A * (-b) * e(-bt) * {2-bt}

d2/dt2[A*t*e^(-bt)] = -Ab(2-bt)e(-bt)

3. Originally Posted by Subhotosh Khan
Thatis incorrect.

d/dt[A*t*e^(-bt)] = A * d/dt[t * e^(-bt)] = A * [t * (-b) * e^(-bt) + e(-bt)] = A * e^(-bt) * [1 - bt]
Thanks for your reply. Sorry if this comes across as unappreciative or rude, but shouldn't that be "e^(-bt)", making my original answer correct?

4. Originally Posted by Whitey
Thanks for your reply. Sorry if this comes across as unappreciative or rude, but shouldn't that be "e^(-bt)", making my original answer correct?
In your answer, A is multiplied with only the first term - that is incorrect.

By the way, if you notice the next line of my response you would see that I had corrected that typo.

5. Originally Posted by Subhotosh Khan
In your answer, A is multiplied with only the first term - that is incorrect.

By the way, if you notice the next line of my response you would see that I had corrected that typo.
That was my mistake. I was meant to write A*((e^(-bt))-(bte^(-bt))), but missed a set of parentheses. Downside to writing functions late at night I guess.

I was more so wondering how you got the (1 - bt) in your first derivative, as this is the only difference to my answer, and I can't quite put it together in my head.

6. Originally Posted by Whitey
That was my mistake. I was meant to write A*((e^(-bt))-(bte^(-bt))), but missed a set of parentheses. Downside to writing functions late at night I guess.

I was more so wondering how you got the (1 - bt) in your first derivative, as this is the only difference to my answer, and I can't quite put it together in my head.
Maybe easier to see without all the paren: let u = e^(-bt), then
A*((e^(-bt))-(bte^(-bt))) = A ( u - b t u) = A u (1 - b t) = A e-bt (1 - b t)
as Subhotosh Khan has.

7. Originally Posted by Ishuda
Maybe easier to see without all the paren: let u = e^(-bt), then
A*((e^(-bt))-(bte^(-bt))) = A ( u - b t u) = A u (1 - b t) = A e-bt (1 - b t)
as Subhotosh Khan has.
So if I am reading this right, (-bte^(-bt)) can be simplified to (1 - bt)? I wasn't aware that this was possible, though the booklet I have been given for this course is quite brief on explanations. Is this done by taking the derivative of (-bte^(-bt))? Or is e able to equal 1 without adding Ln to both sides?

8. Originally Posted by Whitey
So if I am reading this right, (-bte^(-bt)) can be simplified to (1 - bt)? I wasn't aware that this was possible, though the booklet I have been given for this course is quite brief on explanations. Is this done by taking the derivative of (-bte^(-bt))? Or is e able to equal 1 without adding Ln to both sides?
You are not reading it right.

9. Let's simplify the problem:
Originally Posted by Whitey
y=Ate^(-bt)
find t when y'' = 0
We can also drop the A since that is going to come out in the wash so we now have
y = te-bt
by chain rule: (y = f(x)g(x) -> y' = f'(x)g(x) + f(x)g'(x)
y' = e-bt - bt*e-bt
y'' = -b*e-bt - b[e-bt - bt*e-bt]
collect some terms:
y'' = b2t * e-bt - 2b*e-bt
y'' = (b2t -2b)*e-bt
Now solve:
b2t = 2b (since ex != 0 for any x)
t = 2/b