Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
ddx[cos⁡−1 ⁣(xx−(1−x)(1−x2))]\dfrac{d}{dx}\left[\cos^{-1}\!\left(x\sqrt{x}-\sqrt{(1-x)(1-x^{2})}\right)\right] is
A11−x2+12x−x2\dfrac{1}{\sqrt{1-x^{2}}}+\dfrac{1}{2\sqrt{x-x^{2}}}
B−11−x2−12x−x2\dfrac{-1}{\sqrt{1-x^{2}}}-\dfrac{1}{2\sqrt{x-x^{2}}}correct
C11−x2+1x−x2\dfrac{1}{\sqrt{1-x^{2}}}+\dfrac{1}{\sqrt{x-x^{2}}}
D11−x2\dfrac{1}{\sqrt{1-x^{2}}}
Solution
Step 1: Let x=cos⁡A\sqrt{x}=\cos A so A=cos⁡−1 ⁣xA=\cos^{-1}\!\sqrt{x}, and x=cos⁡Bx=\cos B so B=cos⁡−1xB=\cos^{-1}x. Then 1−(x)2=sin⁡A\sqrt{1-(\sqrt{x})^{2}}=\sin A and 1−x2=sin⁡B\sqrt{1-x^{2}}=\sin B. Step 2: The expression inside cos⁡−1\cos^{-1} becomes:
xx−(1−x)(1−x2)=cos⁡B⋅cos⁡A−sin⁡A⋅sin⁡B=cos⁡(A+B)x\sqrt{x}-\sqrt{(1-x)(1-x^{2})}=\cos B\cdot\cos A-\sin A\cdot\sin B=\cos(A+B)
 Step 3: Therefore:
y=cos⁡−1(cos⁡(A+B))=A+B=cos⁡−1 ⁣x+cos⁡−1xy=\cos^{-1}(\cos(A+B))=A+B=\cos^{-1}\!\sqrt{x}+\cos^{-1}x
 Step 4: Differentiate each term. For cos⁡−1 ⁣x\cos^{-1}\!\sqrt{x}: use chain rule with x\sqrt{x}:
ddxcos⁡−1 ⁣x=−11−(x)2⋅12x=−12x 1−x=−12x−x2\dfrac{d}{dx}\cos^{-1}\!\sqrt{x}=\dfrac{-1}{\sqrt{1-(\sqrt{x})^{2}}}\cdot\dfrac{1}{2\sqrt{x}}=\dfrac{-1}{2\sqrt{x}\,\sqrt{1-x}}=\dfrac{-1}{2\sqrt{x-x^{2}}}
 For cos⁡−1x\cos^{-1}x:
ddxcos⁡−1x=−11−x2\dfrac{d}{dx}\cos^{-1}x=\dfrac{-1}{\sqrt{1-x^{2}}}
 Step 5: Combine:
dydx=−12x−x2−11−x2\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{x-x^{2}}}-\dfrac{1}{\sqrt{1-x^{2}}}
 Correct answer: (2)
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