Differentiation and its Application: If x^2+y^2=1 , then | IIT-JEE 2000 Screening

IIT-JEE 2000 ScreeningQuestion 1

If $x^{2}+y^{2}=1$, then

Differentiation and its Application

$yy''-2(y')^{2}+1=0$$yy''+(y')^{2}+1=0$✓ Correct answer$yy''+(y')^{2}-1=0$$yy''+2(y')^{2}+1=0$

Solution

Concept. Differentiate $x^2+y^2=1$ twice implicitly. Formulas used. (1) $\dfrac{d}{dx}(yy')=y'^2+yy''$Step 1 - Once. $x+yy'=0$. Step 2 - Twice. By (1): $1+y'^2+yy''=0$. Answer: (B)

Differentiation and its Application - JEE Mathematics question (IIT-JEE 2000 Screening)

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