How is manipulation of derivatives possible? In parametric differentiation, finding the second derivative can be given by:

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*$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{dy}{dt}*\frac{dt}{dx})$
Now if I am differentiating two functions multiplied together, I would use the product rule right?

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*$\frac{d}{dx}(\frac{dy}{dt}*\frac{dt}{dx})$ = $\frac{dt}{dx}*\frac{d}{dx}(\frac{dy}{dt})$ + $\frac{dy}{dt}* \frac{d}{dx}(\frac{dt}{dx})$
But that is a wrong manipulation, rather this is what is given in my textbook:

Manipulating derivatives is hard, at some points, they act like fractions, while at others they do not. How do I know for sure that what I am doing is correct?
 A: The notation commonly used for higher-order derivatives SHOULD NOT be used as fractions.  In order to treat higher-order derivatives as fractions, the notation needs to be derived from the quotient rule.  For instance, if you want to be able to manipulate the differentials as a fraction, the notation for the second derivative of $y$ with respect to $x$ should be:
$$ \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
You can find that by using the quotient rule on $\frac{dy}{dx}$ and simplifying.
More information about this notation can be found in a paper I wrote titled "Extending the Algebraic Manipulability of Differentials".  The standard notation is suggestive only, and cannot be manipulated as if it were a quotient of actual terms.
Now, we can find the answer to this particular problem without diving too deep into new notation.  A good halfway point is to treat $\frac{d}{dx}\left(\text{expression}\right)$ as actually being $\frac{d(\text{expression})}{dx}$.  So, given that, we can rewrite the formula like this:
$$\frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{d\left(\frac{dy}{dx}\right)}{dt} \times \frac{dt}{dx} $$
Since these are all first order differentials, the terms can cancel, and it is clear why the equivalency exists.
A: What makes you think your first formula was wrong?
For example, let $t = \cos x,$ and let $y = t^2 = \cos^2 x.$ Then
\begin{align}
\frac{dt}{dx} &= -\sin x, \\
\frac{dy}{dt} &= 2t = 2\cos x, \\
\frac{dy}{dx} &= - 2 \sin x\cos x, \\
\frac{d^2y}{dx^2} &= 2 \sin^2 x - 2 \cos^2 x.
\end{align}
Now you can write
$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$ by the chain rule,
and then (as you wrote) the product rule indeed says
\begin{align}
\frac{d}{dx}\left(\frac{dy}{dx}\right)
&= \frac{d}{dx}\left(\frac{dy}{dt} \cdot \frac{dt}{dx}\right) \\
&= \frac{dt}{dx} \cdot \frac{d}{dx}\left(\frac{dy}{dt}\right)
  + \frac{dy}{dt} \cdot \frac{d}{dx}\left(\frac{dt}{dx}\right) \\
&= \left(-\sin x\right) \frac{d}{dx}\left(2\cos x\right)
  + \left(2\cos x\right) \frac{d}{dx}\left(-\sin x\right) \\
&= \left(-\sin x\right) \left(- 2 \sin x\right)
  + \left(2\cos x\right) \left(-\cos x\right) \\
&=  2 \sin^2 x - 2\cos^2 x,
\end{align}
which is the correct result.
What the textbook is doing, on the other hand, is applying the chain rule to the outermost derivative instead of the innermost derivative. That is, they say
$$ \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx} $$
where
$$ u = \frac{dy}{dx}. $$
That's all. It is a simple formula derived by a simple method.
But the fact that there is such a simple formula for rewriting a second derivative does not mean that every other rewriting of the second derivative is wrong.
A: I think what you did and what the textbook did are both right, it's just that you used the chain rule on the inner $\frac{d}{dx}$ to write it as $\frac{dt}{dx}\frac{d}{dt}$, and the textbook used it on the outer $\frac{d}{dx}$, (ie. when writing $\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{d}{dx}y$)
