What Is the Correct Derivative of the Function y(r) = (r^2 - 5.50r)exp(-r)?

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Homework Help Overview

The problem involves finding the first derivative of the function y(r) = (r^2 - 5.50r)exp(-r) at the point r = 5.50. The context is calculus, specifically focusing on differentiation and the application of the chain rule.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of the term exp(-r) and its equivalence to e^(-r). There are attempts to apply the chain rule for differentiation, with some participants expressing confusion about their calculations and results when substituting r = 5.50.

Discussion Status

The discussion is ongoing, with participants sharing their attempts at differentiation and questioning their results. Some guidance has been offered regarding the interpretation of the exponential function, and there is an acknowledgment of the need to revisit calculus concepts.

Contextual Notes

Participants mention uncertainty about their calculus skills and express confusion regarding the correct application of differentiation techniques. There is a reference to needing to consult calculus notes for clarification.

justin335
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Homework Statement


Consider the following function of the variable r, r>/=0
y(r)=(r^(2)-5.50r)exp(-r)
Find the value of the first derivative dy/dr at r=5.50

Homework Equations


How do I solve this? I know its a simple derivative equation, but I can't seem to get it. I tried finding the derivative and then plugging in r=5.50 but MasteringPhysics says it incorrect.
Also, does exp(-r) mean to the -r power? I've never seen it written like that.


The Attempt at a Solution


Assuming exp(-r) means to the negative r power...
I tried using the chain rule;
y(r)=(r^(2)-5.50r)exp(-r)
dy/dr=-r(r^(2)-5.50r)^(-r-1)*(2r-5.50)
plugging in r=5.50, I get a 0, making the solution 0, which is obviously wrong...
I think its my poor calc skills, any help would be great, thanks!
 
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exp(-r) is a representation of ##e^{-r}##.

So, your function is ##y(r) = (r^2-5.50r)e^{-r}##
 
TSny said:
exp(-r) is a representation of ##e^{-r}##.

So, your function is ##y(r) = (r^2-5.50r)e^{-r}##

Thanks, would the derivative be e^-r(2r^3-16.5r^2+30.35r)? Thats what I got, and after subbing in r=5.5 again, I still get the answer wrong...
 
justin335 said:
Thanks, would the derivative be e^-r(2r^3-16.5r^2+30.35r)? Thats what I got, and after subbing in r=5.5 again, I still get the answer wrong...

What's the derivative of ex with respect to x?

What's the derivative of e-x with respect to x?

Do you know how to differentiate by parts?
 
Chestermiller said:
What's the derivative of ex with respect to x?

What's the derivative of e-x with respect to x?

Do you know how to differentiate by parts?

Oh ok, Ill have to reference my calc notes since I am pretty rusty. but I think I should get it now. Thanks for the help
 

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