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Q.Find the derivative of cos⁡x\cos x from first principle. OR Evaluate lim⁡x→01+x−1x\displaystyle\lim_{x\to 0} \dfrac{\sqrt{1+x}-1}{x}.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2026Subjective· 3mImportance★★★★★
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By first principles, f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h} applied to f(x)=cos⁡xf(x)=\cos x gives −sin⁡x-\sin x.

Let f(x)=cos⁡xf(x)=\cos x. By definition: f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \lim_{h\to0}\dfrac{\cos(x+h)-\cos x}{h}.

Using the sum-to-product identity cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A-\cos B = -2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right) with A=x+hA=x+h, B=xB=x:

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

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