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

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2025Subjective· 4mImportance★★★★★
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By definition, the derivative is the limit of the difference quotient; using the sine-difference (sum-to-product) identity turns the limit into a standard sin⁡hh→1\dfrac{\sin h}{h}\to1 form.

Let f(x)=sin⁡2xf(x)=\sin2x. By first principles:

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

Using sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A-\sin B = 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right) with A=2x+2h, B=2xA=2x+2h,\ B=2x:

sin⁡(2x+2h)−sin⁡2x=2cos⁡(2x+h)sin⁡h\sin(2x+2h)-\sin2x = 2\cos(2x+h)\sin h

So:

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