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Q.Find the derivative of the function sin⁡2x\sin 2x from the first Principle.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2018Subjective· 4mImportance★★★★★
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From first principles, ddxsin⁡2x=2cos⁡2x\dfrac{d}{dx}\sin 2x=2\cos 2x.

By definition, with f(x)=sin⁡2xf(x)=\sin 2x,

f′(x)=lim⁡h→0sin⁡2(x+h)−sin⁡2xh=lim⁡h→0sin⁡(2x+2h)−sin⁡2xh.f'(x)=\lim_{h\to0}\frac{\sin 2(x+h)-\sin 2x}{h}=\lim_{h\to0}\frac{\sin(2x+2h)-\sin 2x}{h}.

Use sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2} with C=2x+2h, D=2xC=2x+2h,\ D=2x: …

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