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seen Apr 30 at 11:25

Feb
6
awarded  Scholar
Feb
6
accepted Why is $f'(x) < f(x)/x$ for $f''(x)<0$
Dec
17
comment Why is $f'(x) < f(x)/x$ for $f''(x)<0$
Sorry, I edited the question to bring out more clearly that they are meant to hold for all $x>0$. Better?
Dec
17
revised Why is $f'(x) < f(x)/x$ for $f''(x)<0$
Clarified question by making x>0 an overall requirement.
Dec
17
comment Why is $f'(x) < f(x)/x$ for $f''(x)<0$
Thank you so much, but can you explicitly state the link between the result that $F(x)$ is decreasing and $f'(x) < \frac{f(x)}{x}$? I also don't seem to understand how to get from one to the other.
Dec
16
revised Why is $f'(x) < f(x)/x$ for $f''(x)<0$
Fixed incorrect inequality sign in the title.
Dec
16
asked Why is $f'(x) < f(x)/x$ for $f''(x)<0$
Nov
4
comment Substituting total derivative d for partial derivative \partial
Would this be a correct interpretation of your answer: "When we took the total derivative of the firm's first-order condition we should have written $\partial \tau$ and $\partial k$ instead of $d \tau$ and $d k$, because we were holding all other variables constant. The notational problem wouldn't have appeared in the first place." If this is so, why was I trained to use $d$ instead of $\partial$ in total derivatives and under which conditions should I actually use $d$?
Nov
4
comment Substituting total derivative d for partial derivative \partial
Thank you for taking the time to look into the model. Your explanation somehow resonates well with me, but I still don't understand the 'rule' that is at work here. Could you say a little more as to why the assumption of being in an equilibrium equalizes $\frac{\partial k}{\partial \tau}$ and $\frac{d k}{d \tau}$? How does the economic intuition behind these terms differ outside of an equilibrium?
Oct
21
awarded  Student
Oct
21
awarded  Editor
Oct
21
revised Substituting total derivative d for partial derivative \partial
fixed grammar
Oct
21
asked Substituting total derivative d for partial derivative \partial