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Question Number 36703 by saikiran last updated on 04/Jun/18

show that f(x)=sin X is derivable at every aεR

$${show}\:{that}\:{f}\left({x}\right)=\mathrm{sin}\:{X}\:{is}\:{derivable}\:{at}\:{every}\:{a}\varepsilon{R} \\ $$

Commented by abdo mathsup 649 cc last updated on 04/Jun/18

we have for all x ∈R lim_(h→0) ((f(x+h)−f(x))/h) =lim_(h→0) ((sin(x+h)−sinx)/h) =lim_(h→0) ((sinx cosh +cosx sinh −sinx)/h) =lim_(h→0) (( −sinx(1−cosh))/h) +((sinh)/h) cosx but we know that lim_(h→0) ((1−cosh)/h^2 ) =(1/2) ⇒ lim_(h→0) ((1−cosh)/h) =0 also lim_(h→0) ((sinh)/h) =1 so lim_(h→0) ((f(x+h)−f(x))/h) = cosx (∈R) so f is derivablabe at every point of R .

$${we}\:{have}\:{for}\:{all}\:{x}\:\in{R} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}}\:={lim}_{{h}\rightarrow\mathrm{0}} \frac{{sin}\left({x}+{h}\right)−{sinx}}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{sinx}\:{cosh}\:+{cosx}\:{sinh}\:−{sinx}}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{\:−{sinx}\left(\mathrm{1}−{cosh}\right)}{{h}}\:+\frac{{sinh}}{{h}}\:{cosx}\:{but} \\ $$$${we}\:{know}\:{that}\:{lim}_{{h}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cosh}}{{h}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cosh}}{{h}}\:=\mathrm{0}\:{also}\:{lim}_{{h}\rightarrow\mathrm{0}} \frac{{sinh}}{{h}}\:=\mathrm{1}\:{so} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}}\:=\:{cosx}\:\:\left(\in{R}\right)\:{so}\:{f}\:{is} \\ $$$${derivablabe}\:{at}\:{every}\:{point}\:{of}\:{R}\:. \\ $$

Commented by MJS last updated on 04/Jun/18

to show lim_(h→0) ((sin(x+h)−sin(x−h))/(2h))=cos x we can use sin==((e^(ix) −e^(−ix) )/(2i)); cos x=((e^(ix) +e^(−ix) )/2) lim_(h→0) ((e^(i(x+h)) −e^(−i(x+h)) −e^(i(x−h)) +e^(−i(x−h)) )/(4hi))= =lim_(h→0) (((d/dh)(e^(i(x+h)) −e^(−i(x+h)) −e^(i(x−h)) +e^(−i(x−h)) ))/((d/dh)(4hi)))= =lim_(h→0) (i/(4i))(e^(i(x+h)) +e^(−i(x+h)) +e^(i(x−h)) +e^(−i(x−h)) )= =(1/4)(2e^(ix) +2e^(−ix) )=((e^(ix) +e^(−ix) )/2)=cos x

$$\mathrm{to}\:\mathrm{show} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}+{h}\right)−\mathrm{sin}\left({x}−{h}\right)}{\mathrm{2}{h}}=\mathrm{cos}\:{x} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{sin}==\frac{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2i}};\:\mathrm{cos}\:{x}=\frac{\mathrm{e}^{\mathrm{i}{x}} +\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{i}\left({x}+{h}\right)} −\mathrm{e}^{−\mathrm{i}\left({x}+{h}\right)} −\mathrm{e}^{\mathrm{i}\left({x}−{h}\right)} +\mathrm{e}^{−\mathrm{i}\left({x}−{h}\right)} }{\mathrm{4}{h}\mathrm{i}}= \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{d}}{{dh}}\left(\mathrm{e}^{\mathrm{i}\left({x}+{h}\right)} −\mathrm{e}^{−\mathrm{i}\left({x}+{h}\right)} −\mathrm{e}^{\mathrm{i}\left({x}−{h}\right)} +\mathrm{e}^{−\mathrm{i}\left({x}−{h}\right)} \right)}{\frac{{d}}{{dh}}\left(\mathrm{4}{h}\mathrm{i}\right)}= \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{i}}{\mathrm{4i}}\left(\mathrm{e}^{\mathrm{i}\left({x}+{h}\right)} +\mathrm{e}^{−\mathrm{i}\left({x}+{h}\right)} +\mathrm{e}^{\mathrm{i}\left({x}−{h}\right)} +\mathrm{e}^{−\mathrm{i}\left({x}−{h}\right)} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2e}^{\mathrm{i}{x}} +\mathrm{2e}^{−\mathrm{i}{x}} \right)=\frac{\mathrm{e}^{\mathrm{i}{x}} +\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2}}=\mathrm{cos}\:{x} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

1)meaning of derivative is the ability to draw tangent at a point on a curve. 2)sinx is a continuous curve at every point and tangent can be drawn at everyoint.so it is derivable

$$\left.\mathrm{1}\right){meaning}\:{of}\:{derivative}\:{is}\:{the}\:{ability}\:{to}\:{draw} \\ $$$${tangent}\:{at}\:{a}\:{point}\:{on}\:{a}\:{curve}. \\ $$$$\left.\mathrm{2}\right){sinx}\:{is}\:{a}\:{continuous}\:{curve}\:{at}\:{every}\:{point} \\ $$$${and}\:{tangent}\:{can}\:{be}\:{drawn}\:{at}\:{everyoint}.{so}\:{it} \\ $$$${is}\:{derivable} \\ $$$$ \\ $$

Answered by MJS last updated on 04/Jun/18

f(x)=sin x f′(x)=cos x tangent in P= ((p),((sin p)) ): y=xcos p+sin p −pcos p exists for all p∈R

$${f}\left({x}\right)=\mathrm{sin}\:{x} \\ $$$${f}'\left({x}\right)=\mathrm{cos}\:{x} \\ $$$$\mathrm{tangent}\:\mathrm{in}\:{P}=\begin{pmatrix}{{p}}\\{\mathrm{sin}\:{p}}\end{pmatrix}: \\ $$$${y}={x}\mathrm{cos}\:{p}+\mathrm{sin}\:{p}\:−{p}\mathrm{cos}\:{p} \\ $$$$\mathrm{exists}\:\mathrm{for}\:\mathrm{all}\:{p}\in\mathbb{R} \\ $$

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