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Exercise 10.3 · Q30

Q.Differentiate the following: y=sin⁡−1 ⁣(1−x21+x2)y = \sin^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)

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Step 1. Let u=1−x21+x2u=\dfrac{1-x^2}{1+x^2}, so y=sin⁡−1uy=\sin^{-1}u, and dydu=11−u2\dfrac{dy}{du}=\dfrac{1}{\sqrt{1-u^2}}.

Step 2. Differentiate uu with the quotient rule: dudx=(−2x)(1+x2)−(1−x2)(2x)(1+x2)2=−2x−2x3−2x+2x3(1+x2)2=−4x(1+x2)2\dfrac{du}{dx}=\dfrac{(-2x)(1+x^2)-(1-x^2)(2x)}{(1+x^2)^2}=\dfrac{-2x-2x^3-2x+2x^3}{(1+x^2)^2}=\dfrac{-4x}{(1+x^2)^2}.

Step 3. Compute 1−u21-u^2: 1−u2=(1+x2)2−(1−x2)2(1+x2)21-u^2=\dfrac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}. Using a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b) with a=1+x2, b=1−x2a=1+x^2,\,b=1-x^2: (1+x2)2−(1−x2)2=(2x2)(2)=4x2(1+x^2)^2-(1-x^2)^2=(2x^2)(2)=4x^2. So 1−u2=4x2(1+x2)21-u^2=\dfrac{4x^2}{(1+x^2)^2}.

Step 4. Take the square root (needs ∣x∣|x|, since x2=∣x∣\sqrt{x^2}=|x|): 1−u2=2∣x∣1+x2\sqrt{1-u^2}=\dfrac{2|x|}{1+x^2}.

Step 5. Combine via the chain rule: dydx=du/dx1−u2=−4x/(1+x2)22∣x∣/(1+x2)=−4x2∣x∣(1+x2)=−2x∣x∣(1+x2)\dfrac{dy}{dx}=\dfrac{du/dx}{\sqrt{1-u^2}}=\dfrac{-4x/(1+x^2)^2}{2|x|/(1+x^2)}=\dfrac{-4x}{2|x|(1+x^2)}=\dfrac{-2x}{|x|(1+x^2)}. …

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