Proof of the Chain Rule

\(\text{Rule: If }f\text{ is differentiable at }x\text{ and }g\text{ is differentiable at }y=f(x)\text{,}\)

then g∘f is differentiable at x and (g∘f)′(x)=g′(f(x))f′(x)=g′(y)f′(x).\text{then }g∘f\text{ is differentiable at }x\text{ and }(g∘f)^\prime(x)=g^\prime(f(x))f^\prime(x)=g^\prime(y)f^\prime(x)\text{.}

Preparation:\text{Preparation:}

lim⁡h~→0g(y+h~)−g(y)h~=g′(y)g(y+h~)−g(y)h~=g′(y)+σ(h~), where lim⁡h~→0σ(h~)=0g(y+h~)−g(y)=h~[g′(y)+σ(h~)]\begin{aligned} \lim_{\widetilde{h}\rightarrow0}\frac{g\left(y+\widetilde{h}\right)-g(y)}{\widetilde{h}}&=g^\prime(y)\\ \frac{g\left(y+\widetilde{h}\right)-g(y)}{\widetilde{h}}&=g^\prime(y)+\sigma\left(\widetilde{h}\right),\text{ where }\lim_{\widetilde{h}\rightarrow0}\sigma\left(\widetilde{h}\right)=0\\ g\left(y+\widetilde{h}\right)-g(y)&=\widetilde{h}\left[g^\prime(y)+\sigma\left(\widetilde{h}\right)\right] \end{aligned}

Proof:\text{Proof:}

(g∘f)′(x):=lim⁡h→0(g∘f)(x+h)−(g∘f)(x)h=lim⁡h→0g[f(x+h)]−g[f(x)]h=lim⁡h→0g[f(x)+f(x+h)−f(x)]−g[f(x)]h, Let h~=f(x+h)−f(x)=lim⁡h→0g(y+h~)−g(y)h=lim⁡h→0h~[g′(y)+σ(h~)]h=lim⁡h→0h~hg′(y)+lim⁡h→0h~hσ(h~)=lim⁡h→0f(x+h)−f(x)hg′(y)+lim⁡h→0f(x+h)−f(x)hσ[f(x+h)−f(x)]=g′(y)lim⁡h→0f(x+h)−f(x)h+σ[f(x)−f(x)]lim⁡h→0f(x+h)−f(x)h=g′(y)f′(x)+σ(0)f′(x)=g′(y)f′(x)+0f′(x)(g∘f)′(x)=g′(y)f′(x)■\begin{aligned} (g∘f)^\prime(x)&:=\lim_{h\rightarrow0}\frac{(g∘f)(x+h)-(g∘f)(x)}{h}\\ &=\lim_{h\rightarrow0}\frac{g[f(x+h)]-g[f(x)]}{h}\\ &=\lim_{h\rightarrow0}\frac{g[f(x)+f(x+h)-f(x)]-g[f(x)]}{h},\text{ Let }\widetilde{h}=f(x+h)-f(x)\\ &=\lim_{h\rightarrow0}\frac{g\left(y+\widetilde{h}\right)-g(y)}{h}\\ &=\lim_{h\rightarrow0}\frac{\widetilde{h}\left[g^\prime(y)+\sigma\left(\widetilde{h}\right)\right]}{h}\\ &=\lim_{h\rightarrow0}\frac{\widetilde{h}}{h}g^\prime(y)+\lim_{h\rightarrow0}\frac{\widetilde{h}}{h}\sigma\left(\widetilde{h}\right)\\ &=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}g^\prime(y)+\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\sigma[f(x+h)-f(x)]\\ &=g^\prime(y)\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}+\sigma[f(x)-f(x)]\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\ &=g^\prime(y)f^\prime(x)+\sigma(0)f^\prime(x)\\ &=g^\prime(y)f^\prime(x)+0f^\prime(x)\\ (g∘f)^\prime(x)&=g^\prime(y)f^\prime(x)\quad\blacksquare \end{aligned}

#Calculus

Note by Gandoff Tan
1 year, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Wonderful amazing proof

Sonali Mate - 1 year, 7 months ago
×

Problem Loading...

Note Loading...

Set Loading...