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Решение.

Производная функции, заданной параметрически .

[tex]\left\{\begin{array}{l}\bf x(t)=\dfrac{t}{t+1}\\\bf y(t)=\dfrac{t^2}{3t+4}\end{array}\right\qquad \ \ \bf y'_{x}=\dfrac{y'_{t}}{x'_{t}}\\\\\\x'_{t}=\dfrac{t+1-t}{(t+1)^2}=\dfrac{1}{(t+1)^2}\\\\\\y'_{t}=\dfrac{2t\, (3t+4)-t^2\cdot 3 }{(3t+4)^2}=\dfrac{3t^2+8t}{(3t+4)^2}\\\\\\y'_{x}=\dfrac{y'_{t}}{x'_{t}}=\dfrac{(3t^2+8t)(t+1)^2}{(3t+4)^2}[/tex]    

Вторая производная - это производная от первой производной .

[tex]\bf y''_{xx}=\dfrac{(y'_{x})'_{t}}{x'_{t}}\\\\\\y'_{x}=\dfrac{(3t^2+8t)(t^2+2t+1)}{(3t+4)^2}=\dfrac{3t^4+14t^3+19t^2+8t}{(3t+4)^2}\\\\\\(y'_{x})'_{t}=\dfrac{(12t^3+42t^2+38t+8)(3t+4)^2-(3t^4+14t^3+19t^2+8t)\cdot 6(3t+4)}{(3t+4)^4}=[/tex]

[tex]\bf =\dfrac{(12t^3+42t^2+38t+8)(3t+4)-6(3t^4+14t^3+19t^2+8t)}{(3t+4)^3}=[/tex]  

[tex]\bf =\dfrac{18t^4+90t^3+168t^2+128t+32}{(3t+4)^3}[/tex]

[tex]\bf y''_{xx}=\dfrac{(18t^4+90t^3+168t^2+128t+32)(t+1)^2}{(3t+4)^3}[/tex]