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 Feb6 awarded Scholar Feb6 accepted Why is $f'(x) < f(x)/x$ for $f''(x)<0$ Dec17 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? Dec17 revised Why is $f'(x) < f(x)/x$ for $f''(x)<0$ Clarified question by making x>0 an overall requirement. Dec17 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. Dec16 revised Why is $f'(x) < f(x)/x$ for $f''(x)<0$ Fixed incorrect inequality sign in the title. Dec16 asked Why is $f'(x) < f(x)/x$ for $f''(x)<0$ Nov4 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 $d$? Nov4 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? Oct21 awarded Student Oct21 awarded Editor Oct21 revised Substituting total derivative d for partial derivative \partial fixed grammar Oct21 asked Substituting total derivative d for partial derivative \partial