Second derivatives Find d2ydx2. x+y2=1

Step 1
Given equation is ${\displaystyle x+y^2=1}$.
To find second derivatives ${\displaystyle \frac{d^{2}y}{{dx}^2}}$.
Solution:
Power rule of differentiation states that:
${\displaystyle \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}}$
Chain rule of differentiation states that:
${\displaystyle \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)}$
Step 2
Differentiating the given function with respect to x,
${\displaystyle \frac{d}{dx}(x+y^2)=\frac{d}{dx}\left(1\right)}$
${\displaystyle 1+2y\frac{dy}{dx}=0}$
${\displaystyle 2y\frac{dy}{dx}=-1}$
${\displaystyle \frac{dy}{dx}=-\frac{1}{2y}}$
Again differentiating with respect to x,
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}(-\frac{1}{2y})}$
${\displaystyle =-\frac{1}{2}\frac{d}{dx}\left(\frac{1}{y}\right)}$
${\displaystyle =-\frac{1}{2}\cdot \frac{-1}{y^2}\frac{dy}{dx}}$
${\displaystyle =\frac{1}{2y^2}\cdot \frac{1}{-2y}}$
${\displaystyle =-\frac{1}{4y^3}[U\sin g\frac{dy}{dx}=-\frac{1}{2y}]}$
Step 3
Hence, ${\displaystyle \frac{d^{2}y}{{dx}^2}=-\frac{1}{4y^3}}$.