Question Number 59006 by arcana last updated on 03/May/19
$${probar}\:{con}\:{h}\neq\mathrm{0} \\ $$$$\frac{{sin}\left({x}+{h}\right)−{sin}\left({x}\right)}{{h}}=\frac{{sin}\left({h}/\mathrm{2}\right)}{{h}/\mathrm{2}}{cos}\left({x}+\frac{{h}}{\mathrm{2}}\right) \\ $$$$ \\ $$
Commented by arcana last updated on 25/May/19
![usando 2sin(x)cos(y)=sin(x+y)−sin(y−x) tomar x=(h/2),y=x+(h/2) luego 2sin(h/2)cos(x+h/2)=sin(x+h)−sin(x) (2/h)sin(h/2)cos(x+h/2)=(1/h)[sin(x+h)−sin(x)] ((sin(h/2)cos(x+h/2))/(h/2))=((sin(x+h)−sin(x))/h) ((d sen)/dx)(x) =lim_(h→0) ((sin(x+h)−sin(x))/h)=lim_(h→0) ((sin(h/2)cos(x+h/2))/(h/2)) ((d sen)/dx)(x)=lim_(h→0) ((sin(h/2))/(h/2))lim_(h→0) cos(x+h/2) ((d sen)/dx)(x)=lim_(h→0) cos(x+h/2)=cos(x)](Q60792.png)
$${usando}\:\:\mathrm{2}{sin}\left({x}\right){cos}\left({y}\right)={sin}\left({x}+{y}\right)−{sin}\left({y}−{x}\right) \\ $$$${tomar}\:{x}=\frac{{h}}{\mathrm{2}},{y}={x}+\frac{{h}}{\mathrm{2}} \\ $$$${luego}\:\mathrm{2}{sin}\left({h}/\mathrm{2}\right){cos}\left({x}+{h}/\mathrm{2}\right)={sin}\left({x}+{h}\right)−{sin}\left({x}\right) \\ $$$$\frac{\mathrm{2}}{{h}}{sin}\left({h}/\mathrm{2}\right){cos}\left({x}+{h}/\mathrm{2}\right)=\frac{\mathrm{1}}{{h}}\left[{sin}\left({x}+{h}\right)−{sin}\left({x}\right)\right] \\ $$$$\frac{{sin}\left({h}/\mathrm{2}\right){cos}\left({x}+{h}/\mathrm{2}\right)}{{h}/\mathrm{2}}=\frac{{sin}\left({x}+{h}\right)−{sin}\left({x}\right)}{{h}} \\ $$$$\frac{{d}\:{sen}}{{dx}}\left({x}\right)\:=\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({x}+{h}\right)−{sin}\left({x}\right)}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({h}/\mathrm{2}\right){cos}\left({x}+{h}/\mathrm{2}\right)}{{h}/\mathrm{2}} \\ $$$$\frac{{d}\:{sen}}{{dx}}\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({h}/\mathrm{2}\right)}{{h}/\mathrm{2}}\underset{{h}\rightarrow\mathrm{0}} {{lim}}\:{cos}\left({x}+{h}/\mathrm{2}\right) \\ $$$$\frac{{d}\:{sen}}{{dx}}\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {{lim}cos}\left({x}+{h}/\mathrm{2}\right)={cos}\left({x}\right) \\ $$$$ \\ $$$$ \\ $$