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
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.
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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.
edited Jul 6 "11 in ~ 2:32
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$$\beginalign*\frac\textd^2y\textdx^2&=\frac\textd\textdx\left(\frac\textdy\textdx\right)\\ &=\frac\textd\textdt\left(\frac\textdy\textdx\right).\frac\textdt\textdx=\fracddt\left(\fracdydx\right)\cdot \frac1\fracdxdt\\\endalign*$$where the last equality is together a result of applying the chain rule.
edited Feb 22 "18 in ~ 14:43
answered Jul 6 "11 in ~ 2:45
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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.
answered Jul 6 "11 in ~ 2:26
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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!
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$$
$$ \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$$