Let $\displaystyle x(t)=2t^2+2$ and $\displaystyle y(t)=3t^4+4t^3$, then
find $\displaystyle \frac{d^2y}{dx^2}$ at point (8,80).
Thanks.
To find $\displaystyle \frac{d^2y}{dx^2}$, you must know y(x).
$\displaystyle x=2t^2+2 \Leftrightarrow t=\pm\sqrt{\frac{x-2}{2}}$.
Therefore, $\displaystyle y=3t^4+4t^3 \Leftrightarrow y=3(\frac{x-2}{2})^2\pm 4(\frac{x-2}{2})^{\frac{3}{2}}$
$\displaystyle y(x)=3(\frac{x-2}{2})^2\pm 4(\frac{x-2}{2})^{\frac{3}{2}}$
$\displaystyle y'(x)=?$
$\displaystyle y''(x)=?$