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Exercise 10.4 · Q17

Q.Find the derivative of the following: sin⁡−1(3x−4x3)\sin^{-1}(3x-4x^3)

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Step 1. Let x=sin⁡θx=\sin\theta. Recall the triple-angle identity:

sin⁡3θ=3sin⁡θ−4sin⁡3θ=3x−4x3\sin 3\theta = 3\sin\theta - 4\sin^3\theta = 3x - 4x^3

Step 2. So y=sin⁡−1(3x−4x3)=sin⁡−1(sin⁡3θ)=3θ=3sin⁡−1xy = \sin^{-1}(3x-4x^3) = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x (for ∣x∣≤12|x|\le \tfrac12). …

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