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?
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$\begingroup$To clarify, I assume you mean $$x(t)=5+t^2$$ and $$y(t)=t^2+t^3$$ You can solve for $\frac{d^2y}{dx^2}$ by using the chain rule. To see how to do this, we can re-solve $\frac{dy}{dx}$. \begin{align*} \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}&=\frac{dy}{dt}\left(\frac{dx}{dt}\right)^{-1} \\&=(2t+3t^2)(2t)^{-1} \\&=1+\frac{3t}{2} \end{align*} as you have previously solved.
We can do the same for $\frac{d^2y}{dx^2}$: \begin{align*} \frac{d^2y}{dx^2}&=\frac{d}{dx}\left(\frac{dy}{dx}\right) \\&=\frac{d(\frac{dy}{dx})}{dt}\frac{dt}{dx} \mbox{ (chain rule)} \\&=\frac{d(1+\frac{3t}{2})}{dt}\left(\frac{dx}{dt}\right)^{-1} \\&=\frac{3}{2}(2t)^{-1} \\&=\frac{3}{4t} \end{align*} which is what you wanted. At the end of the day, these sorts of problems boil down to the chain rule, which helps write derivatives in terms of variables you know how to deal with, like $t$.
$\endgroup$ $\begingroup$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:
$$\frac{d}{dx}\frac{dy}{dx} = \frac{d}{dt}\left(1 + \frac{3t}{2}\right)\frac{dt}{dx} = \frac{3}{2}\frac{1}{2t} = \frac{3}{4t}.$$
$\endgroup$ $\begingroup$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 $\frac{d^2y}{dx^2}= \frac{\frac{3}{2}}{2t}= \frac{3t}{4}$
$\endgroup$ 1 $\begingroup$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
$y=\left(\sqrt{x-5}\right)^2+\left(\sqrt{x-5}\right)^3$
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