If $x^2+y^2=1$, then:

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

If $x^2+y^2=1$, then:

Options:

$y.y''-2(y')^2+1=0$

$y.y''+(y')^2+1=0$

$y.y''-(y')^2-1=0$

$y.y''+2(y')^2+1=0$

Correct Answer:

$y.y''+(y')^2+1=0$

Explanation:

We have, $x^2+y^2=1$

$⇒2x+2yy'=0$

$⇒x+yy'=0$

Now, on again differentiating w.r.t. x, we get :

$1+yy''+y'.y'=0$ $∴yy''+(y')^2+1=0$