Find velocity vector from position vector

I'm reviewing for my calculus exam on Tuesday and am a bit stuck on this problem. I must be differentiating something wrong.

**The position vector of a particle is r(t). Find the requested vector.**

Find the velocity vector at time (t)=0,

$\displaystyle r(t)=ln(t^{3}-2t^{2}+3)i - (\sqrt{t^{2}+9})j-8cos(t)k$

Derivative of i component:

$\displaystyle (\frac{3t^{2}-4t}{t^{3}-2t^{2}+3})i$

Derivative of j component:

$\displaystyle (\frac{t}{\sqrt{t^{2}+9}})j$

Derivative of j component:

$\displaystyle 8sin(t)k$

$\displaystyle v(t)=(\frac{3t^{2}-4t}{t^{3}-2t^{2}+3})i$-$\displaystyle (\frac{t}{\sqrt{t^{2}+9}})j$+$\displaystyle 8sin(t)k$

Shouldn't v(0)=0 ?

What am I doing wrong?