To simplify writing and without loss of generality, consider the derivatives of $1/v$ at $t=0$. Your condition on the derivatives of $v$ implies that $v$ can be analytically extended to the open disk of radius $r$ centered at $0$. Moreover, 
$$|v(z)-v(0)|\le\sum_{k=1}^\infty\frac{|v^{(k)}(0)}{k!}\,|z|^k
\le c\sum_{k=1}^\infty\Big(\frac{|z|}r\Big)^k=c\frac{|z|}{r-|z|}\le l/2$$
and hence 
$$|v(z)|\ge l/2$$
for complex $z$ such that  
$$|z|\le s:=\frac l{2c+l}\,r. \tag{1}\label{1}$$

So, for $u:=1/v$ we have $|u(z)|\le 2/l$ given \eqref{1}. So, by the [Cauchy integral formula][1], 
$$|u^{(k)}(0)|\le \frac2l\frac{k!}{s^k} $$
for $k=0,1,\dots$.


  [1]: https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#Theorem