if x=t^(2)+1 and y=t^3 than d^2y/dt^2=?
I was thinking the same as there was an equation given for both
$\displaystyle y=f(t)$ and $\displaystyle x=f(t)$
Therefore a solution could be
$\displaystyle \frac{d^2y}{dx^2} = \frac{\tfrac{d^2y}{dt^2}}{\tfrac{d^2x}{dt^2}}$
$\displaystyle = \frac{6t}{2}$
$\displaystyle = 3t$
We need to be careful...
$\displaystyle x=t^2+1$ $\displaystyle y=t^3$
so $\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{3t^2}{2t}=\frac{3}{2}t$
Now $\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\frac{dy}{dx}} {\frac{dx}{dt}}=\frac{\frac{3}{2}}{2t}=\frac{3}{4t }$
Note we can write y as a function of x as follows
$\displaystyle y=(x-1)^{\frac{3}{2}}$
Now if we take two derivatives we get
$\displaystyle \frac{dy}{dx}=\frac{3}{2}(x-1)^{\frac{1}{2}}$
$\displaystyle \frac{d^2y}{dx^2}=\frac{3}{4}\frac{1}{(x-1)^{\frac{1}{2}}}$
Note that since $\displaystyle x=t^2+1$ we get
$\displaystyle \frac{d^2y}{dx^2}=\frac{3}{4}\frac{1}{(x-1)^{\frac{1}{2}}}=\frac{3}{4}\frac{1}{(t^2+1-1)^{\frac{1}{2}}}=\frac{3}{4t}$