Finding $d^2y/dx^2$ A while ago I did a problem where I needed to find $\frac{d^2y}{dx^2}$ with $x=5+t^2$ and $y=t^2+t^3$. I found $dy/dx$ to be $1+\frac{3t}{2}$. But I never wrote down how I found  $\frac{d^2y}{dx^2}$. Or any other problems like it. The answer was $\frac{3}{4t}$. But I can't figure out how to get to that again. Can someone explain to me how you get to that?
 A: You have to calc. $y(x)$. Therefor $t=\frac{x-5}{2}$
$$\Rightarrow y=\left(\frac{x-5}{2}\right)^2+\left(\frac{x-5}{2}\right)^3$$
Then apply the standard calculations.
If you meant $x=5+t^2$ it's
$y=\left(\sqrt{x-5}\right)^2+\left(\sqrt{x-5}\right)^3$
A: You (likely) used the chain rule to get $dy/dx$:
$$\frac{dx}{dt} = 2t; \frac{dy}{dt} = 2t + 3t^2$$
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 1 + \frac{3t}{2}.$$
Then just use it again:
$$\frac{d}{dx}\frac{dy}{dx} = \frac{d}{dt}\left(1 + \frac{3t}{2}\right)\frac{dt}{dx} = \frac{3}{2}\frac{1}{2t} = \frac{3}{4t}.$$
A: I'm not sure what you mean by "I never wrote down how I found $\frac{d^2y}{dx^2}$.  Do you mean you found It but did not write it down?  Unfortunately I often have the same problem!  You say that "x= 5+ t2".  Did you mean $x= 5+ t^2$?   
Assuming that, $\frac{dx}{dt}= 2t$ and $\frac{dy}{dt}= 2t+ 3t^2$.  So that $\frac{dy}{dx}= \frac{2t+ 3t^2}{2t}= 1+ (3/2)t$.  To find the second derivative, do exactly the same thing again, differentiating the first derivative with respect to x.  Let $Y'= 1+ (3/2)t$, $\frac{d^2y}{dx^2}= \frac{dY'}{dx}= \frac{\frac{dY'}{dt}}{\frac{dx}{dt}}$
$\frac{dY'}{dt}= \frac{3}{2}/4$ and $\frac{dx}{dt}= 2t$ so $\frac{d^2y}{dx^2}= \frac{\frac{3}{2}}{2t}= \frac{3t}{4}$
