Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question
If y=11+x+x2+x3y=\dfrac{1}{1+x+x^{2}+x^{3}}, then (dydx)x=0\left(\dfrac{dy}{dx}\right)_{x=0} and (d2ydx2)x=0\left(\dfrac{d^{2}y}{dx^{2}}\right)_{x=0}
A−1,  0-1,\;0correct
B1,  01,\;0
C0,  20,\;2
D0,  10,\;1
Solution
Step 1: Note y−1=1+x+x2+x3y^{-1}=1+x+x^{2}+x^{3}. At x=0x=0: y−1=1  ⇒  y=1y^{-1}=1\;\Rightarrow\;y=1. Step 2: Differentiate y−1=1+x+x2+x3y^{-1}=1+x+x^{2}+x^{3} implicitly:
−y−2dydx=1+2x+3x2  ⇒  dydx=−y2(1+2x+3x2)-y^{-2}\dfrac{dy}{dx}=1+2x+3x^{2}\;\Rightarrow\;\dfrac{dy}{dx}=-y^{2}(1+2x+3x^{2})
 At x=0x=0, y=1y=1:
dydx∣x=0=−1\dfrac{dy}{dx}\bigg|_{x=0}=-1
 Step 3: Differentiate dydx=−y2(1+2x+3x2)\dfrac{dy}{dx}=-y^{2}(1+2x+3x^{2}):
d2ydx2=− ⁣[2ydydx(1+2x+3x2)+y2(2+6x)]\dfrac{d^{2}y}{dx^{2}}=-\!\left[2y\dfrac{dy}{dx}(1+2x+3x^{2})+y^{2}(2+6x)\right]
 At x=0x=0, y=1y=1, dydx=−1\dfrac{dy}{dx}=-1:
d2ydx2∣x=0=− ⁣[2(1)(−1)(1)+12(2)]=−[−2+2]=0\dfrac{d^{2}y}{dx^{2}}\bigg|_{x=0}=-\!\left[2(1)(-1)(1)+1^{2}(2)\right]=-[-2+2]=0
 Correct answer: (1) −1,  0-1,\;0
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