What is the derivative of u(w)=k-e^(-aw)

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SUMMARY

The derivative of the function u(w) = k - e^(-aw) is correctly calculated as u'(w) = ae^(-aw). The discussion highlights the application of the chain rule in differentiation, particularly for exponential functions. Additionally, the second question regarding the inequality U(w) = sqrt(w) is addressed, emphasizing the proof of the convexity property of the square root function. The participants clarify the notation and confirm the correctness of the derivative calculation.

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Homework Statement



I have two questions. First:

What is the derivative of u(w)=k-e^(-aw)

Second:

For U(w)=sqrt(w), prove that U(pie(x)+(1-pie)y) > pie*U(x)+(1-pie)U(y)

Homework Equations





The Attempt at a Solution



For the first one, I haven't seen this type of problem in so long that I just don't remember how to take a derivative in this form. Would it be:

u'(w)= -(-aw)e^(-aw-1)*-a
=(aw)e^(a^2w+a)?

I wasn't sure if the chain rule was appropriate here.

For the second question, I have:

sqrt(pie(x)+(1-pie)y) > pie*sqrt(x)+(1-pie)sqrt(y) so...

(pie(x)+(1-pie)y)^(1/2) > pie*(x)^1/2+(1-pie)(y)^1/2

From here I don't know where to go. I don't have much experience with proofs so can anyone give some guidance? Thanks for your help!
 
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e^(aw) is not a power law. d/dw(exp(aw)) is a*exp(aw). Try again.
 


Sorry I'm a little confused by your reply, does that mean for U(w)=k-e^(-aw) that
U'(w)=(k-e^(-aw))' =-a(-e)^(-aw) =ae^(-aw)? Thanks again for your help.
 


3.141592654 said:
Sorry I'm a little confused by your reply, does that mean for U(w)=k-e^(-aw) that
U'(w)=(k-e^(-aw))' =-a(-e)^(-aw) =ae^(-aw)? Thanks again for your help.

The answer is right. -a(-e)^(-aw) is a bit confusing. You might want to move a parenthesis in there.
 


hmm I see the notation is a bit weird. But the answer comes out correct if written as
-a(-e^(-aw)) right? Thanks again.
 


Sure. I just didn't like (-e)^(-aw).
 

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