If $F(x)$ is the role with parameter eliminated then $\displaystyle F'(x) = \frac\textdy\textdt\big/\frac\textdx\textdt$

But the procedure for taking the 2nd derivative is just defined as " replace $y$ through dy/dx " to obtain

$$\frac\textd^2y\textdx^2=\frac\textd\textdx\left(\frac\textdy\textdx\right)=\frac\left<\frac\textd\textdt\left(\frac\textdy\textdt\right)\right>\left(\frac\textdx\textdt\right)$$

I don"t recognize the justification because that this step. No at all.

But that"s all my publication says ~ above the issue then that launches in to plugging points in come this formula, and it appears to work well enough, but I don"t understand why.

You are watching: D^2y/dx^2 parametric

I often find answers about question ~ above differentials are past my level, I"d really prefer to get this, it"d mean a lot come me if someone could break that down.

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edited Jul 6 "11 in ~ 2:32

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request Jul 6 "11 in ~ 2:14

NoteventhetutorknowsNoteventhetutorknows

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## 4 answers 4

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Consider

$$\beginalign*\frac\textd^2y\textdx^2&=\frac\textd\textdx\left(\frac\textdy\textdx\right)\\ &=\frac\textd\textdt\left(\frac\textdy\textdx\right).\frac\textdt\textdx=\fracddt\left(\fracdydx\right)\cdot \frac1\fracdxdt\\\endalign*$$where the last equality is together a result of applying the chain rule.

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edited Feb 22 "18 in ~ 14:43

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answered Jul 6 "11 in ~ 2:45

NanaNana

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Their justification is the you deserve to use the same process for $\fracdydx$ together for $Y$ since you deserve to now consider $Y_2 = g_2(t) = \fracdydx(t)$, the is, you as soon as again have actually a parametric equation in terms of the parameter $t$, and the parametric equation because that $x$ continues to be the same.

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answered Jul 6 "11 in ~ 2:26

VhailorVhailor

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I"m not solid enough to understand when you can treat dy/dx together a "fraction" and when you can"t. But the method I reasoned it was as follows...

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dy/dx = dy/dt * dt/dx (very chain rule-esque)

d2y/dx2 = d2y/dxdt * dt/dx

Which seems entirely continuous with Nana"s answer!

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reply Oct 29 "16 in ~ 8:28

grandfather PicklesMr Pickles

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I"ve come up through a "simple enough" proof for the parametric derivative. Start by think about a curve parameterized in the means :$$, then I will rewrite derivative operator v $x$ :

$$ \fracdy(\theta)dx = \fracd \thetadx \fracd yd \theta$$The over equality hold by the chain rule, derivative that inner duty times derivative of outer function, currently I will remove the $y$:

$$ \fracddx = \fracd \thetadx \fracd d \theta \tag1$$

Now, the above can be believed of together an operator which eats a function and maps to another function, for instance if us multiply both side on ideal by $y$ , then it provides the $ \fracdydx$ through the parametric derivative the $y$.

And, ns will present one an ext identity:

$$ \theta( x( \theta) ) = \theta \tag2$$

The over can be believed of in the complying with way: The role $x(\theta)$ maps native $ \theta \to x$ , the duty $ \theta(x)$ maps indigenous $ x \to \theta$, merely the inverse mapping and hence over equality hold true for at the very least a little interval of where we can say the $ \theta $ and also $x$ are bijective functions of each other. Taking the derivate that $(2)$ through $ \theta$,:

$$ \frac d \thetadx \fracdxd \theta = 1$$

Or,

$$ \fracd\thetadx = \frac1 \frac d xd \theta \tag2$$

Hence, we deserve to rewrite (1) as:

$$ \fracddx = \frac \fracd d \theta \fracdxd \theta \tag3$$

Now, main point $y$ on ideal side the (3):$$ \fracdydx = \frac \fracdy d \theta \fracdxd \theta \tag4$$