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Question Number 11581 by Nayon last updated on 28/Mar/17

cAn you prove the chAin role??

$${cAn}\:{you}\:{prove}\:{the}\:{chAin}\:{role}?? \\ $$

Answered by linkelly0615 last updated on 28/Mar/17

yes∼

$${yes}\sim \\ $$

Commented by linkelly0615 last updated on 28/Mar/17

(BELOW)

$$\left({BELOW}\right) \\ $$

Answered by linkelly0615 last updated on 28/Mar/17

  set that [f(g(x))]′ exist    [f(g(x))]′=lim_(h→0) (([f(g(x+h))]−[f(g(x))])/h)  ∵g(x+h)=g(x)+hg′(x)+(h^2 /2)g′′(x)+......  ∴when “h” is small enough  then  g(x+h)≈g(x)+hg′(x)  ⇒[f(g(x+h))]≈[f(g(x)+hg′(x))]  if “hg′(x)” is small enough  then  [f(g(x)+hg′(x))]≈[f(g(x))+(hg′(x))f′(g(x))]  ...  ⇒  [f(g(x))]′=lim_(h→0) (([f(g(x+h))]−[f(g(x))])/h)  =lim_(h→0) (([f(g(x)+hg′(x))]−[f(g(x))])/h)  =lim_(h→0) (([f(g(x))+(hg′(x))f′(g(x))]−[f(g(x))])/h)  =f′(g(x))g′(x)

$$ \\ $$$${set}\:{that}\:\left[{f}\left({g}\left({x}\right)\right)\right]'\:{exist} \\ $$$$ \\ $$$$\left[{f}\left({g}\left({x}\right)\right)\right]'=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[{f}\left({g}\left({x}+{h}\right)\right)\right]−\left[{f}\left({g}\left({x}\right)\right)\right]}{{h}} \\ $$$$\because{g}\left({x}+{h}\right)={g}\left({x}\right)+{hg}'\left({x}\right)+\frac{{h}^{\mathrm{2}} }{\mathrm{2}}{g}''\left({x}\right)+...... \\ $$$$\therefore{when}\:``{h}''\:{is}\:{small}\:{enough} \\ $$$${then} \\ $$$${g}\left({x}+{h}\right)\approx{g}\left({x}\right)+{hg}'\left({x}\right) \\ $$$$\Rightarrow\left[{f}\left({g}\left({x}+{h}\right)\right)\right]\approx\left[{f}\left({g}\left({x}\right)+{hg}'\left({x}\right)\right)\right] \\ $$$${if}\:``{hg}'\left({x}\right)''\:{is}\:{small}\:{enough} \\ $$$${then} \\ $$$$\left[{f}\left({g}\left({x}\right)+{hg}'\left({x}\right)\right)\right]\approx\left[{f}\left({g}\left({x}\right)\right)+\left({hg}'\left({x}\right)\right){f}'\left({g}\left({x}\right)\right)\right] \\ $$$$... \\ $$$$\Rightarrow \\ $$$$\left[{f}\left({g}\left({x}\right)\right)\right]'=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[{f}\left({g}\left({x}+{h}\right)\right)\right]−\left[{f}\left({g}\left({x}\right)\right)\right]}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[{f}\left({g}\left({x}\right)+{hg}'\left({x}\right)\right)\right]−\left[{f}\left({g}\left({x}\right)\right)\right]}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[{f}\left({g}\left({x}\right)\right)+\left({hg}'\left({x}\right)\right){f}'\left({g}\left({x}\right)\right)\right]−\left[{f}\left({g}\left({x}\right)\right)\right]}{{h}} \\ $$$$={f}'\left({g}\left({x}\right)\right){g}'\left({x}\right)\:\: \\ $$

Commented by linkelly0615 last updated on 28/Mar/17

I think it might be what you want.

$${I}\:{think}\:{it}\:{might}\:{be}\:{what}\:{you}\:{want}. \\ $$

Commented by Nayon last updated on 29/Mar/17

Can you prove it any easy way   without using any series?

$${Can}\:{you}\:{prove}\:{it}\:{any}\:{easy}\:{way}\: \\ $$$${without}\:{using}\:{any}\:{series}? \\ $$

Commented by linkelly0615 last updated on 29/Mar/17

Commented by linkelly0615 last updated on 29/Mar/17

Maybe it is okay  if not ...I will try...  actually, I am trying to find   some diffrent ways to prove   that rule ...  (Including a crazy way)

$${Maybe}\:{it}\:{is}\:{okay} \\ $$$${if}\:{not}\:...{I}\:{will}\:{try}... \\ $$$$\mathrm{actually},\:{I}\:{am}\:{trying}\:{to}\:{find}\: \\ $$$${some}\:{diffrent}\:{ways}\:{to}\:{prove}\: \\ $$$${that}\:{rule}\:... \\ $$$$\left({Including}\:{a}\:{crazy}\:{way}\right) \\ $$$$ \\ $$

Commented by Nayon last updated on 29/Mar/17

it is not seen properlly.... please  comment a clear one..

$${it}\:{is}\:{not}\:{seen}\:{properlly}....\:{please} \\ $$$${comment}\:{a}\:{clear}\:{one}.. \\ $$

Commented by linkelly0615 last updated on 29/Mar/17

it can zoom in by your finger...

$${it}\:{can}\:{zoom}\:{in}\:{by}\:{your}\:{finger}... \\ $$$$ \\ $$

Commented by linkelly0615 last updated on 29/Mar/17

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