Найти dy/dx и d2y/dx2

[latex] \left \{ {{x=t^4} \atop {y=t^2+t}} \right. \\\\ 1)\quad \frac{dy}{dx} = \frac{y'_{t}}{x'_{t}} = \frac{2t+1}{4t^3} \\\\2)\quad \frac{d^2y}{dx^2} = \frac{\left (\frac{2t+1}{4t^3}\right )'}{x'_{t}} = \frac{ \frac{2\cdot 4t^3-12t^2(2t+1)}{16t^9} }{4t^3} = \frac{-16t^3-12t^2}{64t^{12}} =- \frac{4t+3}{16t^{10}} [/latex]