Find the derivative of a complicated expression

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SUMMARY

The discussion focuses on finding the derivative of the function P(t) = 2048000(t^2)(e^1600t) + 5120t(e^-1600t) + 3.2(e^-1600t). The correct approach involves applying the product rule of differentiation in parts, specifically using functions f(t) = 2048000(t^2), g(t) = e^1600t, h(t) = 5120t, and p(t) = e^-1600t. The final derivative is expressed as dP/dt = [f(t)g(t)]' + [h(t)p(t)]' + 3.2[e^-1600t]'. The user expresses uncertainty about their calculations, but the method outlined is confirmed as valid.

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  • Review the product rule of differentiation in calculus.
  • Practice differentiating exponential functions, particularly those involving products.
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Homework Statement


Find dP/dt of 2048000(t^2)(e^1600t)+5120t(e^-1600t)+3.2(e^-1600t)


Homework Equations





The Attempt at a Solution


2048000(2t(e^-1600t)-1600(t^2)(e^-1600t))+5120((e^-1600t)-1600(e^-1600t))+3.2(-1600(e^-1600t))
After moving the t over to simply the first part and multiplying thru I get:
4096000t-3276800000(t^2)+5120-8192000-5120(e^-1600t)

Can this be correct? For some reason it seems like I did something wrong, but I keep checking the numbers and can't figure out what it could be.

Thanks in advance!
 
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The way you have presented your question is rather confusing so I am going to assume a function P(t) is equal to 2048000(t^2)(e^1600t)+5120t(e^-1600t)+3.2(e^-1600t). The trick to differentiating these kind of 'functions' is to really do it in parts, at least until you're comfortable using all the rules of differentiation. For example, try letting f(t) = 2048000(t^2), and g(t) = (e^1600t), then determine [f(t)g(t)]`. Next, let h(t) = 5120t, and p(t) = e^-1600t, then determine [h(t)p(t)]`. And of course, at last, take the derivative of the last term. So your derivative dP/dt = [f(t)g(t)]` + [h(t)p(t)]` + 3.2[e^-1600t]`. You can do this because the linearity of the derivative operation. At any rate, make sense?
 

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