Question

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

• A. $yy"-2{\left(y\text{'}\right)}^2+1=0$
• B. $yy"+{\left(y\text{'}\right)}^2+1=0$
• C. $yy"-{\left(y\text{'}\right)}^2+1=0$
• D. None of these

Answer 1

The correct option is B

$yy"+{\left(y\text{'}\right)}^2+1=0$

Step $1:$ Differentiate the given equation.

Differentiate $x^2+y^2=1$ with respect to $x$.

$\Rightarrow 2x+2yy\text{'}=0...\left(1\right)$ [Since, $\frac{d\left(x^n\right)}{dx}=n{x}^{n-1}$]

Step $2:$ Differentiate equation $1$.

Differentiate equation $1$ with respect to $x$.

$$\begin{gathered} 2+2{\left(y\text{'}\right)}^2+2yy"=0 \\ \Rightarrow yy"+{\left(y\text{'}\right)}^2+1=0 \end{gathered}$$ [Since, $\frac{d\left(f\right(x\left)g\right(x\left)\right)}{dx}=f\left(x\right)g\text{'}\left(x\right)+f\text{'}\left(x\right)g\left(x\right)$]

Hence, option $B$ is the correct answer.