# derivative inequality

• Nov 20th 2007, 01:14 AM
miss_lolitta
derivative inequality

I have this problem..

if $\displaystyle x(t)^{'}$ is a nonnegative and decreasing function and

$\displaystyle x(t)$ is a positive and nondecreasing

so..

$\displaystyle \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t}$ for all $\displaystyle t\geqslant b$ ??!

I tried to prove this problem but at this step

$\displaystyle x(t)-x(b)\geqslant(t-b)x(t)^{'}$

I don't know how to get the rest of proof from this step..:confused:

can someone help me to get the rest of proof??

thanks for any help
• Nov 20th 2007, 08:29 AM
ThePerfectHacker
What is $\displaystyle x(t)$, what is $\displaystyle x'(t)$, on what interval are they defined? What is $\displaystyle b$?
• Nov 20th 2007, 11:40 AM
miss_lolitta
moredetial
if$\displaystyle x^{'}(t)$ is a nonnegative and decreasing function
and $\displaystyle x(t)$ is a positive and nondecreasing on [b,t]

then

$\displaystyle x(t)-x(b)=\int^{t}_{b}{x^{'}(s)ds}$

since $\displaystyle x^{'}(t)$ is decreasing

$\displaystyle x^{'}(t)\leqslant x^{'}(s)$

this implies

$\displaystyle x(t)-x(b)\geqslant x^{'}(t) \int^{t}_{b}{ds}$

so

$\displaystyle x(t)-x(b)\geqslant(t-b)x^{'}(t)$

i.e.

$\displaystyle \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t-b}$.....(*)

now,I just want to get

$\displaystyle \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t}$??!

I hope that's clear