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

$$showthatf\left(x\right)=\sin Xisderivableateverya\epsilon R$$

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

$$wehaveforallx\in R$$$${lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}={lim}_{h\to 0}\frac{sin(x+h)-sinx}{h}$$$$={lim}_{h\to 0}\frac{sinxcosh+cosxsinh-sinx}{h}$$$$={lim}_{h\to 0}\frac{-sinx(1-cosh)}{h}+\frac{sinh}{h}cosxbut$$$$weknowthat{lim}_{h\to 0}\frac{1-cosh}{h^2}=\frac{1}{2}\Rightarrow$$$${lim}_{h\to 0}\frac{1-cosh}{h}=0also{lim}_{h\to 0}\frac{sinh}{h}=1so$$$${lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}=cosx(\in R)sofis$$$$derivablabeateverypointofR.$$

Commented by MJS last updated on 04/Jun/18

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

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

$$1)meaningofderivativeistheabilitytodraw$$$$tangentatapointonacurve.$$$$2)sinxisacontinuouscurveateverypoint$$$$andtangentcanbedrawnateveryoint.soit$$$$isderivable$$$$$$

Answered by MJS last updated on 04/Jun/18

$$f\left(x\right)=\sin x$$$$f^{\prime }\left(x\right)=\cos x$$$$\mathrm{tangent}\mathrm{in}P=\left(\begin{array}{c}p\\ \sin p\end{array}\right):$$$$y=x\cos p+\sin p-p\cos p$$$$\mathrm{exists}\mathrm{for}\mathrm{all}p\in \mathbb{R}$$

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