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I'm learning multiple applications of the chain rule and the notation surrounding it. Does the following notation with example correctly reflect the chain rule in both Lagrange and Leibniz notation?

The derivative: $$ y = f(x) $$ $$ \frac {dy}{dx} = f'(x) = \frac {dy}{dx} \Big|_{x} = \lim_{h \to 0} {\frac {f(h+x) - f(x)}{(h+x) -x}} = \lim_{x_1 \to x} {\frac {f(x_1) - f(x)}{x_1 -x}} $$ or when: $$ x=a, \frac {dy}{dx} = f'(a) $$

The chain rule: $$ y = f(u), u =g(x) $$ $$ \frac {dy}{du} \Big|_{g(x)} \cdot \frac {du}{dx} \Big|_{x} = (f \circ g)'(x) = (f' \circ g)(x) \cdot g'(x) $$

The chain rule applied two times: $$ f(x) = (f_1 \circ (f_2 \circ f_3))(x) $$

$$ \begin{align*} f'(x) &= (f_1 \circ (f_2 \circ f_3))'(x) = (f_1' \circ (f_2 \circ f_3))(x) \cdot (f_2 \circ f_3)'(x) \\ &= (f_1' \circ (f_2 \circ f_3))(x) \cdot (f_2' \circ f_3)(x) \cdot f_3'(x) \end{align*} $$

Decompose composite into separate functions: $$ f'(x) = \frac {d f} {dx} = \frac {dh} {dj} \frac {dj} {dk} \frac {dk} {dx} $$ $$ f = h = f_1(j), j = f_2(k), k = f_3(x) $$

$$ f'(x) = \frac {d f} {dx} \Big|_{x} = \frac {dh} {dj} \Big|_{f_2(f_3(x))}\frac {dj} {dk} \Big|_{f_3(x)} \frac {dk} {dx} \Big|_{x} $$

which agrees with above.

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I think that the source of your confusion arises from blurring the distinction between a function and its value. For example, if we write $f(x) = y$, then $x$ and $y$ are numbers and $f$ is the function that maps the number $x$ to the number $y$. In the two notations for the derivative, we write $$ f'(x) = \frac{dy}{dx} $$

Notice that I don't ever write the function name in the Leibniz differential notation. For compositions such as $$ x \overset{g}{\longmapsto} u \overset{f}{\longmapsto} y, $$ the derivative in the two notations are $$ g'(x) = \frac{du}{dx}, \qquad f'(u) = \frac{dy}{du}, $$ and $$ (f \circ g)'(x) = f'\bigl( g(x) \bigr) \, g'(x) = f'(u) \, g'(x) = \frac{dy}{du} \, \frac{du}{dx}. $$

This generalizes naturally to chains of functions of any length, hence the name the chain rule.

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  • $\begingroup$ I updated the OP to clarify what I'm asking. I think you covered some of it. $\endgroup$
    – Nick
    Feb 25, 2021 at 6:05

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