what is going on - A level further maths edexcel pearson question

jas[er

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I'm very confused on this question, not sure if its some differentiation techniques I havent learnt in pure yet like implicit or quotient stuff, below is the question and supposed solution in answers:
 

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What techniques have you learned? As they state, this uses the chain rule and product rule (repeatedly and with complicated expressions), but nothing more.

At what step in the solution are you confused?
 
only very recently learned the chain rule and product rule but never seen a question like this usually only questions in form secx(x^2+1)^3 or something, im really just confused with the whole thing. the question initially gives you d^2/dx^2 (y^2) but usually, id see dy/dx (y), given y or higher derivatives of that... I'm not really sure where to start with the question.
 
only very recently learned the chain rule and product rule but never seen a question like this usually only questions in form secx(x^2+1)^3 or something, im really just confused with the whole thing. the question initially gives you d^2/dx^2 (y^2) but usually, id see dy/dx (y), given y or higher derivatives of that... I'm not really sure where to start with the question.
The solution is already GIVEN to you!

Exactly which step confuses you?
 
What is confusing about this is that it is ABSTRACT. You are used to applying the chain rule to a specific function. Now you are being asked to use the chain rule on a generic function that is not specified.

But obviously y is a function of x, meaning that the value of y is fully determined by x. And likewise

[MATH]\dfrac{dy}{dx}[/MATH] is a function of x. Therefore

[MATH]\text {Define } p(x) = \dfrac{dy}{dx} = f'(x) \implies p'(x) = f''(x) = \dfrac{d^2y}{dx^2}.[/MATH]
[MATH]\text {Define } q(x) = x^2 \implies y^2 = q(y) = q(f(x)) \implies[/MATH]
[MATH]q'(x) = \left ( \dfrac{d}{dy} \ y^2 \right ) * \dfrac{dy}{dx} = 2y * p(x) = 2\{f(x) * p(x)\}.[/MATH]
Simple chain rule. With me so far?

[MATH]q'(x) = 2\{f(x) * p(x)\} \implies q''(x) = 2\{y * p'(x) + f'(x) * p(x)\}.[/MATH]
Simple product rule. Still with me?

But p(x) = f'(x) and p'(x) = f''(x).

[MATH]q''(x) = 2\{y * f''(x) + [f'(x)]^2\} \implies[/MATH]
[MATH]q''(x) = \left ( \dfrac{d^2}{dx^2}\ y^2 \right ) = 2y * \dfrac{d^2y}{dx^2} + 2 * \left ( \dfrac{dy}{dx} \right )^2.[/MATH]
 
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