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Q.If y=(cot⁡−1x3)2y = (\cot^{-1} x^3)^2, find dydx\dfrac{dy}{dx}.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2018Subjective· 2mImportance★★★★★
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Chain rule with the derivative ddxcot⁡−1u=−11+u2dudx\frac{d}{dx}\cot^{-1}u=\frac{-1}{1+u^2}\frac{du}{dx} gives −6x2cot⁡−1(x3)1+x6\dfrac{-6x^2\cot^{-1}(x^3)}{1+x^6}.

Step 1: Set up the chain rule

Let u=cot⁡−1(x3)u=\cot^{-1}(x^3), so y=u2y=u^2 and dydx=2ududx\dfrac{dy}{dx}=2u\dfrac{du}{dx}.

Step 2: Differentiate u

dudx=−11+(x3)2⋅ddx(x3)=−3x21+x6\dfrac{du}{dx} = \dfrac{-1}{1+(x^3)^2}\cdot\dfrac{d}{dx}(x^3) = \dfrac{-3x^2}{1+x^6}

Step 3: Combine …

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