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Q.Find the derivative of sin⁡x\sin x from First Principle.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2026Subjective· 3mImportance★★★★★
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ddx(sin⁡x)=cos⁡x\dfrac{d}{dx}(\sin x) = \cos x.

By the first-principles (limit) definition of the derivative:

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

Use the sum-to-product identity 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=x+hA=x+h, B=xB=x:

sin⁡(x+h)−sin⁡x=2cos⁡(x+h2)sin⁡(h2)\sin(x+h)-\sin x = 2\cos\left(x+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)

So:

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