Proof of the Chain Rule

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

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

$\text{Preparation:}$

\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}

$\text{Proof:}$

\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}

Note by Gandoff Tan
1 year, 9 months ago

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- 1 year, 9 months ago