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Q.Find the derivatives of the following functions w.r.t. xx:

(a) 1x2\dfrac{1}{x^2}
(b) sin⁡2x\sin^2 x OR Find the derivative of f(x)=x+1xf(x) = x + \dfrac{1}{x} from the first principle.
Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2022Subjective· 5mImportance★★★★★
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Use the power rule for (a), the chain rule for (b); for the OR-alternative, apply the first-principles definition f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.

(a) 1x2=x−2\dfrac{1}{x^2}=x^{-2}. By the power rule, ddx(xn)=nxn−1\dfrac{d}{dx}(x^n)=nx^{n-1}:

ddx(x−2)=−2x−3=−2x3\frac{d}{dx}(x^{-2}) = -2x^{-3} = -\frac{2}{x^3}

(b) sin⁡2x=(sin⁡x)2\sin^2x = (\sin x)^2. By the chain rule, ddx[(sin⁡x)2]=2sin⁡x⋅ddx(sin⁡x)=2sin⁡xcos⁡x\dfrac{d}{dx}[(\sin x)^2] = 2\sin x\cdot\dfrac{d}{dx}(\sin x) = 2\sin x\cos x. Using the double-angle identity 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\sin2x:

ddx(sin⁡2x)=sin⁡2x\frac{d}{dx}(\sin^2x) = \sin2x


OR alternative: derivative of f(x)=x+1xf(x)=x+\dfrac1x from first principles.

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(x+h+1x+h)−(x+1x)hf'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{\left(x+h+\dfrac{1}{x+h}\right)-\left(x+\dfrac1x\right)}{h}

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