Proof of the Chain Rule

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

then gf is differentiable at x and (gf)(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:}

limh~0g(y+h~)g(y)h~=g(y)g(y+h~)g(y)h~=g(y)+σ(h~), where limh~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:}

(gf)(x):=limh0(gf)(x+h)(gf)(x)h=limh0g[f(x+h)]g[f(x)]h=limh0g[f(x)+f(x+h)f(x)]g[f(x)]h, Let h~=f(x+h)f(x)=limh0g(y+h~)g(y)h=limh0h~[g(y)+σ(h~)]h=limh0h~hg(y)+limh0h~hσ(h~)=limh0f(x+h)f(x)hg(y)+limh0f(x+h)f(x)hσ[f(x+h)f(x)]=g(y)limh0f(x+h)f(x)h+σ[f(x)f(x)]limh0f(x+h)f(x)h=g(y)f(x)+σ(0)f(x)=g(y)f(x)+0f(x)(gf)(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}

Note by Gordon Chan
1 month, 1 week ago

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Wonderful amazing proof

Sonali Mate - 1 month, 1 week ago

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