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# Differentiation

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1. Can someone help me with this question?

v = 20000te^(-t/3)
Use this model and calculus to find the value of t when dv/dt = 0

I can't work out from the mark scheme what they did to get dv/dt. The final answer should be 3.
Thanks
2. (Original post by dont)
Can someone help me with this question?

v = 20000te^(-t/3)
Use this model and calculus to find the value of t when dv/dt = 0

I can't work out from the mark scheme what they did to get dv/dt. The final answer should be 3.
Thanks
Product rule?
3. (Original post by dont)
Can someone help me with this question?

v = 20000te^(-t/3)
Use this model and calculus to find the value of t when dv/dt = 0

I can't work out from the mark scheme what they did to get dv/dt. The final answer should be 3.
Thanks
As mentioned above, you would use the product rule for this
You should get dv/dt = 2000e^(-t/3) - ((2000/3)t^2)e^(-t/3)
Spoiler:
Show
Sorry for the messy layout I'm on mobile

You are correct that this would need to equal 0, you could try to factorise this to make it a bit nicer which would help you solve it too.
4. (Original post by KaylaB)
As mentioned above, you would use the product rule for this
You should get dv/dt = 2000e^(-t/3) - ((2000/3)t^2)e^(-t/3)
Spoiler:
Show
Sorry for the messy layout I'm on mobile
You are correct that this would need to equal 0, you could try to factorise this to make it a bit nicer which would help you solve it too.
Ah thanks. Could you possible go through the steps you did to get dv/dt as I struggle with the product rule.
5. (Original post by dont)
Ah thanks. Could you possible go through the steps you did to get dv/dt as I struggle with the product rule.
You split the functions in to two parts, usually shown as f(x) = uv where you're able to diffrentiate u and v individually
So f'(x) = (du/dx)v + (dv/dx)u
So basically you diffrentiate one and multiply it by the original of the other. Then do vice versa and add them
So if f(x) = 4xe^2x = (4x)(e^2x) (So u=4x and v=e^2x)
du/dx = 4
dv/dx = 2e^2x
So f'(x) = (4)(e^2x) + (2e^2x)(4x) = 4e^2x + 8xe^2x
= (4e^2x)(1+2x) which is just simplified
6. (Original post by KaylaB)
You split the functions in to two parts, usually shown as f(x) = uv where you're able to diffrentiate u and v individually
So f'(x) = (du/dx)v + (dv/dx)u
So basically you diffrentiate one and multiply it by the original of the other. Then do vice versa and add them
So if f(x) = 4xe^2x = (4x)(e^2x) (So u=4x and v=e^2x)
du/dx = 4
dv/dx = 2e^2x
So f'(x) = (4)(e^2x) + (2e^2x)(4x) = 4e^2x + 8xe^2x
= (4e^2x)(1+2x) which is just simplified
I get it now. Thanks that really helped
7. (Original post by dont)
I get it now. Thanks that really helped
No problem

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