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Question Number 123147 by aurpeyz last updated on 23/Nov/20

from first principle obtain the   diffrential coefficient of cos(x)

$${from}\:{first}\:{principle}\:{obtain}\:{the}\: \\ $$$${diffrential}\:{coefficient}\:{of}\:{cos}\left({x}\right) \\ $$

Answered by TANMAY PANACEA last updated on 23/Nov/20

(dy/dx)= lim_(h→0)  ((f(x+h)−f(x))/h)=lim_(h→0) ((cos(x+h)−cosx)/h)  ((d(cosx))/dx)=lim_(h→0)  ((2sin(x+(h/2)))/((h/2)×2))×(−sin(h/2))  =sin(x+(0/2))×−1=−sinx

$$\frac{{dy}}{{dx}}=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{cos}\left({x}+{h}\right)−{cosx}}{{h}} \\ $$$$\frac{{d}\left({cosx}\right)}{{dx}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{sin}\left({x}+\frac{{h}}{\mathrm{2}}\right)}{\frac{{h}}{\mathrm{2}}×\mathrm{2}}×\left(−{sin}\frac{{h}}{\mathrm{2}}\right) \\ $$$$={sin}\left({x}+\frac{\mathrm{0}}{\mathrm{2}}\right)×−\mathrm{1}=−{sinx} \\ $$

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