Cabiolab

2021-02-22

Second derivatives Find $\frac{{d}^{2}y}{d{x}^{2}}$.
$x+{y}^{2}=1$

Arnold Odonnell

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

Step 3
Hence, $\frac{{d}^{2}y}{{dx}^{2}}=-\frac{1}{4{y}^{3}}$.

Do you have a similar question?