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Q.Find the derivative of cos⁡x\cos x from first principle. OR Compute the derivative of f(x)=sin⁡2xf(x) = \sin^2 x.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2024Subjective· 4mImportance★★★★★
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Applying the first-principle definition f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} with f(x)=cos⁡xf(x)=\cos x, and simplifying via cos⁡C−cos⁡D\cos C-\cos D identity, gives −sin⁡x-\sin x.

Step 1. By definition, f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \displaystyle\lim_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h}.

Step 2. Use cos⁡C−cos⁡D=−2sin⁡(C+D2)sin⁡(C−D2)\cos C - \cos D = -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right) with C=x+hC=x+h, D=xD=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\left(\dfrac{h}{2}\right).

Step 3. So f′(x)=lim⁡h→0−2sin⁡(x+h2)sin⁡(h2)h=lim⁡h→0−sin⁡(x+h2)⋅sin⁡(h/2)h/2f'(x) = \displaystyle\lim_{h\to 0} \dfrac{-2\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} = \displaystyle\lim_{h\to 0} -\sin\left(x+\dfrac{h}{2}\right) \cdot \dfrac{\sin(h/2)}{h/2}.

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