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Q.Compute the derivative of sin⁡2x\sin 2x from the first principle.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2026Subjective· 4mImportance★★★★★
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First principles with sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C-\sin D=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2} gives 2cos⁡2x2\cos2x.

Let f(x)=sin⁡2xf(x)=\sin2x. By definition,

f′(x)=lim⁡h→0sin⁡2(x+h)−sin⁡2xh=lim⁡h→0sin⁡(2x+2h)−sin⁡2xh.f'(x)=\lim_{h\to0}\frac{\sin2(x+h)-\sin2x}{h}=\lim_{h\to0}\frac{\sin(2x+2h)-\sin2x}{h}.

Using sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C-\sin D=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2} with C=2x+2hC=2x+2h, D=2xD=2x: …

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