Method of DifferentiationhardFree

Method of Differentiation: Equals

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

Question
If y=xa2−1−2a2−1tan⁡−1 ⁣(sin⁡xa+a2−1+cos⁡x)y=\dfrac{x}{\sqrt{a^{2}-1}}-\dfrac{2}{\sqrt{a^{2}-1}}\tan^{-1}\!\left(\dfrac{\sin x}{a+\sqrt{a^{2}-1}+\cos x}\right), where a∈(−∞,−1)∪(1,∞)a\in(-\infty,-1)\cup(1,\infty), then y′ ⁣(π2)y'\!\left(\dfrac{\pi}{2}\right) equals
A1a\dfrac{1}{a}correct
B2a\dfrac{2}{a}
C12a\dfrac{1}{2a}
Daa
Solution
Step 1: Let A=a+a2−1A=a+\sqrt{a^{2}-1}. Differentiate yy:
dydx=1a2−1−2a2−1⋅11+(sin⁡xA+cos⁡x)2⋅(A+cos⁡x)cos⁡x+sin⁡2x(A+cos⁡x)2\dfrac{dy}{dx}=\dfrac{1}{\sqrt{a^{2}-1}}-\dfrac{2}{\sqrt{a^{2}-1}}\cdot\dfrac{1}{1+\left(\dfrac{\sin x}{A+\cos x}\right)^{2}}\cdot\dfrac{(A+\cos x)\cos x+\sin^{2}x}{(A+\cos x)^{2}}
 Step 2: The numerator of the inner derivative simplifies to:
ddx ⁣(sin⁡xA+cos⁡x)=Acos⁡x+cos⁡2x+sin⁡2x(A+cos⁡x)2=Acos⁡x+1(A+cos⁡x)2\dfrac{d}{dx}\!\left(\dfrac{\sin x}{A+\cos x}\right)=\dfrac{A\cos x+\cos^{2}x+\sin^{2}x}{(A+\cos x)^{2}}=\dfrac{A\cos x+1}{(A+\cos x)^{2}}
 Step 3: At x=π2x=\dfrac{\pi}{2}: sin⁡π2=1\sin\dfrac{\pi}{2}=1 and cos⁡π2=0\cos\dfrac{\pi}{2}=0. The tan⁡−1\tan^{-1} argument =1A=\dfrac{1}{A}, so 1+(1A)2=A2+1A21+\left(\dfrac{1}{A}\right)^{2}=\dfrac{A^{2}+1}{A^{2}}. The inner derivative at x=π2x=\dfrac{\pi}{2} becomes A⋅0+1A2=1A2\dfrac{A\cdot 0+1}{A^{2}}=\dfrac{1}{A^{2}}. Step 4:
dydx∣x=π/2=1a2−1−2a2−1⋅A2A2+1⋅1A2=1a2−1(1−2A2+1)=1a2−1⋅A2−1A2+1\dfrac{dy}{dx}\bigg|_{x=\pi/2}=\dfrac{1}{\sqrt{a^{2}-1}}-\dfrac{2}{\sqrt{a^{2}-1}}\cdot\dfrac{A^{2}}{A^{2}+1}\cdot\dfrac{1}{A^{2}}=\dfrac{1}{\sqrt{a^{2}-1}}\left(1-\dfrac{2}{A^{2}+1}\right)=\dfrac{1}{\sqrt{a^{2}-1}}\cdot\dfrac{A^{2}-1}{A^{2}+1}
 Step 5: Substitute A=a+a2−1A=a+\sqrt{a^{2}-1}:
A2=2a2−1+2aa2−1,A2−1=2(a2−1)+2aa2−1=2a2−1(a+a2−1)A^{2}=2a^{2}-1+2a\sqrt{a^{2}-1},\quad A^{2}-1=2(a^{2}-1)+2a\sqrt{a^{2}-1}=2\sqrt{a^{2}-1}(a+\sqrt{a^{2}-1})
A2+1=2a2+2aa2−1=2a(a+a2−1)A^{2}+1=2a^{2}+2a\sqrt{a^{2}-1}=2a(a+\sqrt{a^{2}-1})
 So:
A2−1A2+1=2a2−1(a+a2−1)2a(a+a2−1)=a2−1a\dfrac{A^{2}-1}{A^{2}+1}=\dfrac{2\sqrt{a^{2}-1}(a+\sqrt{a^{2}-1})}{2a(a+\sqrt{a^{2}-1})}=\dfrac{\sqrt{a^{2}-1}}{a}
 Step 6: Substituting back:
dydx∣x=π/2=1a2−1⋅a2−1a=1a\dfrac{dy}{dx}\bigg|_{x=\pi/2}=\dfrac{1}{\sqrt{a^{2}-1}}\cdot\dfrac{\sqrt{a^{2}-1}}{a}=\dfrac{1}{a}
 Correct answer: (1)
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