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I am not sure how to do this but I need to find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$

For $x = t^2 + 1, y= t^2+t$

And then show what t values gives a concave upward.

I know the simple formula to find $\frac{dy}{dx}$

I get $\frac{dy}{dx} = \frac{y'}{x'}$ $\frac{dy}{dx} = \frac{2t+1}{2t}$

$\frac{d^2 y}{dx^2} = \frac{\frac{dy}{dx}}{dx}$

$\frac{\frac{2t+1}{2t}}{2t}$

This is wrong and I am not sure why, they end with a negative number which makes no sense to me.

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    Let $f(x)=\mathrm{d}y/\mathrm{d}x = y'(t)/x'(t)$. Then shouldn't $\mathrm{d}f/\mathrm{d}x = f'/x'$ where primes are differentiation with respect to $t$? [It's been many years since I've done this, so I may be wrong...]2012-06-25

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You have $\frac{dy}{dx}=\frac{2t+1}{2t}=1+\frac1{2t}\;.$

To differentiate this again with respect to $x$, you must repeat what you did to get this: calculate

$\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}\;.$

You forgot to do the differentiation in the numerator. When you do it, you get

$\frac{\frac{d}{dt}\left(1+\frac1{2t}\right)}{2t}=\frac{-\frac1{2t^2}}{2t}=-\frac1{4t^3}\;.$

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    @Jordan: That’s just the derivative with respect to $t$ of whatever’s inside the parentheses. In this case it’s the derivative w.r.t. $t$ of $1+\frac1{2t}$.2012-06-25
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Note that $\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dt}(\frac{dy}{dx}).\frac{dt}{dx}=\frac{d}{dt}(\frac{2t+1}{2t}).\frac{1}{2t}=....$