B Proof of Chain Rule: Understanding the Limits

AI Thread Summary
The discussion focuses on the proof of the Chain Rule in calculus, demonstrating how the derivative of a composite function is derived. The function E(k) is defined, and its continuity at k=0 is established, leading to the conclusion that the derivative of f(g(x)) can be expressed as f'(g(x))g'(x). Participants clarify doubts about the limits involving E(k) and the constancy of f'(g(x)) during the limit process, emphasizing that E(k) does not depend on h and that f'(g(x)) remains unchanged as h approaches 0. The conversation concludes with affirmations of understanding regarding these mathematical concepts.
mcastillo356
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I've got a proof of the Chain Rule, but basic questions about basic steps
First I quote the text, and then the attempts to solve the doubts:

"Proof of the Chain Rule

Be ##f## a differentiable function at the point ##u=g(x)##, with ##g## a differentiable function at ##x##. Be the function ##E(k)## described this way:
$$E(0)=0$$
$$E(k)=\dfrac{f(u+k)-f(u)}{k}-f'(u)\qquad\mbox{if}\;k\neq 0$$

By definition of derivative, ##\lim_{k \to{0}}{E(k)}=f'(u)-f'(u)=0=E(0)##, so ##E(k)## is continuous in ##k=0##. Also, be ##k=0## or not, we have

$$f(u+k)-f(u)=(f'(u)+E(k))k$$

Be now ##u=g(x)## and ##k=g(x+h)-g(x)##, so ##u+k=g(x+h)##; we obtain

$$f(g(x+h))-f(g(x))=(f'(g(x))+E(k))(g(x+h)-g(x))$$

As ##g## is differentiable at ##x##, ##\lim_{h \to{0}}{[g(x+h)-g(x)]/h}=g'(x)##. Also, ##g## is continuous at ##x##, by Theorem 1
THEOREM 1 Being differentiable means being continous

If ##f## is differentiable at ##x##, we know it exists

$$\displaystyle\lim_{h \to{0}}{\dfrac{f(x+h)-f(x)}{h}}=f'(x)$$

Using the rules of limits (Theorem 2 of section 1.2)

Rules for limits

If ##\lim_{x \to a}{f(x)}=L##, ##\lim_{x \to a}{g(x)}=M##, and ##k## is a constant, then

1. Limit of a sum: ##\displaystyle\lim_{x \to{a}}{[f(x)+g(x)]}=L+M##
2. Limit of a subtraction: ##\displaystyle\lim_{x \to{a}}{[f(x)+g(x)]}=L+M##
3. Limit of a product: ##\displaystyle\lim_{x \to{a}}{f(x)g(x)}=LM##
4. Limit of a function multiplied by a constant: ##\displaystyle\lim_{x \to{a}}{kf(x)}=kL##
5. Limit of a division: ##\displaystyle\lim_{x \to a}{\dfrac{f(x)}{g(x)}}=\dfrac{L}{M}\qquad\mbox{if}\;M\neq 0##
If ##m## is an integer and ##n## a positive integer, then
6. Limit of a power: ##\displaystyle\lim_{x \to{a}}{[f(x)]^{m/n}}=L^{m/n}##, whenever ##L>0## if ##n## is even, and ##L\neq 0## if ##m<0##
If ##f(x)\geq g(x)## at an interval that contains ##a## inside, then
7. Order preservation: ##L\geq M##

we have

$$\lim_{h \to{0}}{f(x+h)-f(x))}=\lim_{h \to{0}}{\left (\dfrac{f(x+h)-f(x)}{h}\right)(h)}=(f'(x))(0)=0$$

This is equivalent to ##\lim_{h \to{0}}{f(x+h)=f(x)}##, and means ##f## is continous.So ##\lim_{h \to{0}}{E(k)}=\lim_{h \to{0}}{(g(x+h)-g(h)=0}##. As ##E## is continuous in 0, ##\lim_{h \to{0}}{E(k)}=\lim_{k \to{0}}{E(k)}=E(0)=0##. This way,

$$\dfrac{d}{dx}f(g(x))=\displaystyle\lim_{h \to{0}}{\dfrac{f(g(x+h))-f(g(x))}{h}}$$

$$ \qquad\qquad\qquad=\displaystyle\lim_{h \to{0}}{(f'(g(x)+E(k))\dfrac{g(x+h)-g(x)}{h}}$$

$$\qquad\qquad\qquad=(f'(g(x)+0)g(x)=f'(g(x))g'(x)$$

As we wanted to prove."

Doubts:

-¿##\lim_{h \to{0}}{E(k)}=\lim_{k \to{0}}{E(k)}##?

Attempt: ##\lim_{h \to{0}}{E(k)}=\displaystyle\lim_{h \to{0}}{\dfrac{E(k+h)-E(k)}{h}}=\displaystyle\lim_{k \to{0}}{\dfrac{E(k)-E(0)}{k-0}}## (Bad, I guess)

-¿Why ##f'(g(x))## remains the same at the last step?:

$$ \qquad\qquad\qquad=\displaystyle\lim_{h \to{0}}{(f'(g(x)+E(k))\dfrac{g(x+h)-g(x)}{h}}$$

$$\qquad\qquad\qquad=(f'(g(x)+0)g(x)=f'(g(x))g'(x)$$

Attempt: are different variables

Greetings to everybody, have a nice St Joseph's Day!
 
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mcastillo356 said:
-¿limh→0E(k)=limk→0E(k)?
No. ##\lim_{h \to 0} E(k) = E(k)##, since E(k) doesn't involve h in any way.

For the same reason, if f(x) = 2x + 3, ##\lim_{z \to 19} f(x) = f(x)##
 
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Greetings
 
mcastillo356 said:
-¿Why f′(g(x)) remains the same at the last step?:
For the same reason as before -- f'(g(x)) doesn't involve h, so in the limit process (with h approaching 0), f'(g(x)) is unaffected.
 
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