Question Number 129033 by oustmuchiya@gmail.com last updated on 12/Jan/21
$${prove}\:{using}\:{the}\:{first}\:{principle}\:{that} \\ $$$${the}\:{derivative}\:{of}\:\boldsymbol{{sin}}\:\boldsymbol{{x}}\:{is}\:\boldsymbol{{cox}}\:\boldsymbol{{x}}\:{and} \\ $$$${that}\:{the}\:{derivative}\:{of}\:\boldsymbol{{cos}}\:\boldsymbol{{x}}\:{is} \\ $$$$−\boldsymbol{{sinx}} \\ $$
Commented by bramlexs22 last updated on 12/Jan/21
$$\mathrm{y}=\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{y}'\:=\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{h}\right)−\mathrm{sin}\:\mathrm{x}}{\mathrm{h}} \\ $$$$\:\:\:\:\:=\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:\left(\frac{\mathrm{2x}+\mathrm{h}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{h}}{\mathrm{2}}\right)}{\mathrm{h}} \\ $$$$\:\:\:\:=\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{h}/\mathrm{2}\right)}{\mathrm{h}}\:×\:\mathrm{2cos}\:\mathrm{x} \\ $$$$\:=\:\mathrm{cos}\:\mathrm{x}\: \\ $$