Proof of Chain Rule: Understanding the Limits

In summary: This is also why the E(k) term disappears in the last step. In summary, the proof shows that the derivative of a composite function f(g(x)) is equal to the product of the derivatives of f(x) and g(x). The proof uses the rules of limits and the definition of the derivative to show that the limit of a function E(k) is equal to 0. This allows for the simplification of the equation and ultimately proves the desired result.
  • #1
mcastillo356
Gold Member
570
280
TL;DR Summary
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!
 
  • Like
Likes annabrown
Mathematics news on Phys.org
  • #2
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)##
 
  • Love
Likes mcastillo356
  • #3
Thanks!
Greetings
 
  • #4
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.
 
  • Like
Likes mcastillo356

Related to Proof of Chain Rule: Understanding the Limits

1. What is the chain rule in calculus?

The chain rule is a mathematical rule used in calculus to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

2. How is the chain rule used to find limits?

The chain rule is used to find limits by breaking down a complex function into simpler functions and evaluating the limit of each function separately. This allows us to find the limit of the composite function by using the chain rule to combine the limits of the individual functions.

3. What is the difference between the chain rule and the product rule?

The chain rule and the product rule are both rules used in calculus to find derivatives. The main difference is that the chain rule is used for composite functions, where the product rule is used for multiplying two or more functions together.

4. Can the chain rule be applied to any composite function?

Yes, the chain rule can be applied to any composite function, as long as the individual functions are differentiable. This means that the functions must have a well-defined derivative at every point in their domain.

5. How can the chain rule be extended to higher dimensions?

The chain rule can be extended to higher dimensions by using the multivariate chain rule, which is a generalization of the chain rule for functions with multiple variables. It states that the derivative of a multivariate composite function is equal to the sum of the partial derivatives of the outer function with respect to each variable, multiplied by the derivative of the inner function with respect to that same variable.

Similar threads

Replies
4
Views
828
  • General Math
Replies
2
Views
942
  • Calculus and Beyond Homework Help
Replies
12
Views
500
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
345
  • General Math
Replies
11
Views
1K
Replies
4
Views
1K
Replies
1
Views
894
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
392
Back
Top