at t time a particle is moving along the x axis at position x. the relationship between x and t is given by tx=x^2 +8. at x = 2 the velocity of the particle is:
Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.
Then:
$\displaystyle t=x^{2}+8$
then
$\displaystyle x=\pm \sqrt[ ]{ t-8} $
then
$\displaystyle velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}$
Then if x = 2, velocity = 1/4
Seem reasonable?