- #1

mcastillo356

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- 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:

"

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

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)

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

1.

2.

3.

4.

5.

If ##m## is an integer and ##n## a positive integer, then

6.

If ##f(x)\geq g(x)## at an interval that contains ##a## inside, then

7.

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!

"

**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!