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EXERCISE 1.2 · Q63

Q.Differentiate the following w.r.t. xx: sin⁡−1(1+x22)\sin^{-1}\left(\dfrac{1+x^2}{2}\right)

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Let u=1+x22u=\dfrac{1+x^2}{2}, so u′=xu'=x. Using ddxsin⁡−1u=u′1−u2\dfrac{d}{dx}\sin^{-1}u=\dfrac{u'}{\sqrt{1-u^2}}, we need 1−u2=1−(1+x2)24=4−(1+x2)241-u^2=1-\dfrac{(1+x^2)^2}{4}=\dfrac{4-(1+x^2)^2}{4}. Factor the numerator as a difference of squares: 4−(1+x2)2=[2−(1+x2)][2+(1+x2)]=(1−x2)(3+x2)4-(1+x^2)^2=[2-(1+x^2)][2+(1+x^2)]=(1-x^2)(3+x^2). So 1−u2=(1−x2)(3+x2)41-u^2=\dfrac{(1-x^2)(3+x^2)}{4}, and $\sqrt{1-u^2}=\dfrac{\sqrt{(1-x^2)(3+x^2) …

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