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We are often taught that $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$$ Why are we allowed to say this?

What about the case of higher derivaitves, i.e.

$$\frac{\frac{d^ny}{dt^n}}{\frac{d^nx}{dt^n}} $$ Can these be reduced to remove the $dt$s?

(Please mark as duplicate if it is, I'm having trouble finding a similar question)

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4 Answers

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It is much more complex for higher derivatives.

For the second derivative from a parametric system $x(t),y(t)$, we have $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}$$ $$\frac{d}{dt}\left(\frac{{y'}}{{x'}}\right)\frac{1}{{x'}}=\frac{{x'}{y''} - {y'}{x''}}{{x'}^3}$$ where the $'$ and $''$ correspond to the derivatives with respect to $t$.

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It doesn't work nicely for higher derivatives. For example, $$\dfrac{d^2y }{d t^2} = \dfrac{dy}{dx} \dfrac{d^2 x}{d t^2} + \left(\dfrac{dx}{dt} \right)^2 \dfrac{d^2y}{dx^2}$$

The case $n=1$ follows from the chain rule: $$\dfrac{d}{dt} y(x(t)) = y'(x(t)) \times x'(t)$$ which is $$\dfrac{dy}{dx} \dfrac{dx}{dt} $$

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First question

The equatlity comes from the total differential.

Let $z=(x,y)$, then

$dz=\frac{\partial z}{\partial x}\cdot dx +\frac{\partial z}{\partial y}\cdot dy$

Setting $dz=0$

$\frac{\partial z}{\partial x}\cdot dx +\frac{\partial z}{\partial y}\cdot dy=0$

$\frac{\partial z}{\partial x}\cdot dx =-\frac{\partial z}{\partial y}\cdot dy$

$$\Large{\frac{\frac{\partial z}{\partial x}}{\frac{\partial z}{\partial y}}\normalsize{ =-\frac{dy}{dx}}}$$

The expressions for the partial derivatives cannot be splittet or exchanged. In general $\Large{\frac{1}{\frac{\partial z}{\partial y}}}\neq \frac{\partial y}{\partial z}$.

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The equation is only a notational shorthand for the chain rule. You need to distinguish carefully in each derivative what is being seen as a function, and what is being treated as an independent parameter.

In higher derivatives: the derivative say is expressed as a product of the other two. In taking another derivative, by product rule, more terms enter the scene, each of which requires derivation by chain rule. That's why a simple relation does not hold.

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