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Here's what I did:

$\begin{array}{l} y=x^{2} \\ \Rightarrow \frac{d y}{d x}=2 x \end{array}$

and we know that $x=\pm \sqrt{y}$, let's take the $+$ for the sake of the example.

so $\frac{d y}{d x}=2 \sqrt{y}$

so $\begin{array}{c} \frac{d}{d y}\left(\frac{d y}{d x}\right)= \frac{1}{\sqrt{y}} \end{array}$

But on Wolfram Alpha it gives $0$ because it switches automatically from $\frac{d} {dy} $ to $\frac{\partial}{\partial y}$. But still$\begin{array}{c} \frac{\partial }{\partial y}\left(\frac{d y}{d x}\right) \end{array} = \begin{array}{l} \frac{\partial}{\partial y}(2 \sqrt{y}) =\frac{1}{\sqrt{y}} \end{array} $

I'm so confused by when can I take derivatives or partial derivatives.

A. P.
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elie makdissi
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    I am afraid the question may make no sense without context. Classical Leibniz notation does not produce self-contained expressions, but heavily relies on the context. – Alexey Jan 16 '23 at 09:03
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    If you put in $\frac{d}{dy} \big( \frac{dy}{dx} \big)$ without specifying to Wolfram alpha that $y(x) = x^2$ it will return $0$, as there is no assigned value. – Dstarred Jan 16 '23 at 09:08
  • @Alexey - that's pretty much the context, which is to check that if that's true or not. – Elie Makdissi Jan 16 '23 at 10:09
  • @Dstarred - i indeed wrote that $y=x^{2}$ – Elie Makdissi Jan 16 '23 at 10:10
  • @ElieMakdissi, i meant that i am afraid that the given context is insufficient to provide a meaningful answer, but I may be wrong. – Alexey Jan 16 '23 at 12:29
  • The notation $\frac{\partial}{\partial y}$ is especially sensitive to context. It assumes you have a function defined with the syntax $f(\ldots,y,\ldots)$ where each "$\ldots$" represents zero or more other variables. This doesn't play well with variables that already have defined relationships such as $y=x^2.$ – David K Jan 16 '23 at 14:50

3 Answers3

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Assuming $x,y>0,$ it follows that $x=\sqrt y$ so $$\frac{d}{dy}(\frac{dy}{dx})=\frac{d}{dy}(2x)$$ $$=\frac{d}{dy}(2\sqrt y)$$ $$=\frac{1}{\sqrt y}.$$

P. Lawrence
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2

By the chain rule,

$$\frac{d\frac{dy}{dx}}{dy} = \frac{d\frac{dy}{dx}}{dx} \cdot \frac{dx}{dy} = 2 \cdot \frac{\operatorname{sgn}(x)}{2\sqrt y}$$

since $y(x)=x^2\implies x(y)=\pm\sqrt y$.

user170231
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0

When you substitute $x$ for $\sqrt y$, you also need to swap $dx$ for $dy$.

JMP
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