Помогите найти первую и вторую производную

Решите задачу:

$\left \{ {{x=\sqrt{t^2-1}} \atop {y=\frac{t+1}{\sqrt{t^2-1}}}} \right. \\\\x'_{t}=\frac{2t}{2\sqrt{t^2-1}}\; ;\\\\y'_{t}=\frac{\sqrt{t^2-1}-(t+1)\cdot \frac{t}{\sqrt{t^2-1}}}{t^2-1}= \frac{t^2-1-t(t+1)}{(t^2-1)^{\frac{3}{2}}} = \frac{-t-1}{(t^2-1)^{3/2}} \; ;\\\\y'_{x}= \frac{y'_{t}}{x'_{t}}= \frac{-(t+1)\sqrt{t^2-1}}{t\cdot (t^2-1)^{3/2}}=-\frac{t+1}{t\cdot (t^2-1)}=- \frac{t+1}{t^3-t} \; ;\\\\(y'_{x})'_{t}=- \frac{t^3-t-(t+1)(3t^2-1)}{(t^3-t)^2} = -\frac{t^3-t-3t^3+t-3t^2+1}{t^2(t^2-1)^2} =$

$=\frac{2t^3+3t^2-1}{t^2(t^2-1)^2}\; ;\\\\y''_{xx}= \frac{(y'_{x})'_{t}}{x'_{t}}=\frac{(2t^3+3t^2-1)\sqrt{t^2-1}}{t^3(t^2-1)^2} =\frac{2t^3+3t^2-1}{t^3(t^2-1)^{3/2}}\; ;$