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

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


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...] – Alex Nelson Jun 25 '12 at 18:10
up vote 4 down vote accepted

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


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


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I do not understand the d/dt part, what that means and how I calculate that. – user138246 Jun 25 '12 at 18:37
@Jordan: $\frac{d...}{dt}$ is just a notation for what is you looking for. It means you differentiate $(...)$ with respect of $t$. – Babak S. Jun 25 '12 at 18:45
@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 $1+\frac1{2t}$. – Brian M. Scott Jun 25 '12 at 18:46

Note that $$\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dt}(\frac{dy}{dx}).\frac{dt}{dx}=\frac{d}{dt}(\frac{2t+1}{2t}).\frac{1}{2t}=....$$

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