If $x^2+y^2=1$, then:
Answer & explanation
Correct answer: option 2
We have, $x^2+y^2=1$
$⇒2x+2yy'=0$
$⇒x+yy'=0$
Now, on again differentiating w.r.t. x, we get :
$1+yy''+y'.y'=0$ $∴yy''+(y')^2+1=0$
If $x^2+y^2=1$, then:
Correct answer: option 2
We have, $x^2+y^2=1$
$⇒2x+2yy'=0$
$⇒x+yy'=0$
Now, on again differentiating w.r.t. x, we get :
$1+yy''+y'.y'=0$ $∴yy''+(y')^2+1=0$