Hi, I've already proven that dy/dx=1/2{1-(9/[2t^2+1])}
How do I show that dy/dx<1/2??
Thanks a lot!
Hi cyt91,
$\displaystyle \frac{dy}{dx}=\frac{1}{2}\left[1-\left(\frac{9}{2t^2+1}\right)\right]=\frac{1}{2}\left[\frac{2t^2+1}{2t^2+1}-\frac{9}{2t^2+1}\right]$
$\displaystyle =\frac{1}{2}\left[\frac{2t^2+1-9}{2t^2+1}\right]$
and as the numerator of the second factor is clearly less than the denominator, the second factor is less than one,
hence the overall expression is less than 0.5