How do I implicitly differentiate $\frac{d}{dx}(2x(y')^2y'')$

Question

I know that I will probably need to use the 3-way chain rule: $$(fgh)'=f'gh+fg'h+fgh',$$ but since I'm differentiating in terms of $x$, I'm very confused as to how to to "append" $y'$ terms to the end of each term in the product rule expansion. For instance, I believe $\frac{d}{dx}(xy)=y+xy'(y')$, where I have "appended" a $y'$ to the latter term. This makes sense from the (informal) "fraction analogy" in Leibniz notation: $$\frac{d}{dx}=\frac{d}{dy}\cdot\frac{dy}{dx},$$ but I have trouble generalising to more complicated examples like the one mentioned above.

It would be much appreciated if someone can write out the full solution or explain how they got it.

Note: $y=y(x)$ (is this noteworthy?)

Answer 1

Since $$ ((y')^2)' = \frac{\mathrm{d}(y')^2}{\mathrm{d}x} = \frac{\mathrm{d}(y')^2}{\mathrm{d}y'} \frac{\mathrm{d}y'}{\mathrm{d}x} = 2y'y'', $$ your 3-way chain rule leads to $$ \begin{align} (2x(y')^2y'')' &= (2x)'(y')^2y'' + 2x((y')^2)'y'' + 2x(y')^2(y'')' \\ &= 2(y')^2y'' + 4xy'(y'')^2 + 2x(y')^2y'''. \end{align} $$

Answer 2

Use logarithmic differentiation $$f(x)=2x(y')^2y''$$ $$\log(f(x))=\log(2)+\log(x)+2\log(y')+\log(y'')$$ $$\frac{f'(x)}{f(x)} =\frac d {dx} \log(f(x))=\frac 1x +2\frac{y''}{y'}+\frac{y'''}{y''}$$ $$f'(x)=\left( \frac 1x +2\frac{y''}{y'}+\frac{y'''}{y''}\right) f(x)$$

Simplify.