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, $$|u^{(k)}(0)|\le \frac2l\frac{k!}{s^k} $$ for $k=0,1,\dots$.

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, $$|u^{(k)}(0)|\le \frac2l\frac{k!}{s^k} $$ for $k=0,1,\dots$.