Question: "Second derivatives Find d2ydx2. x+y2=1"

Cabiolab

Answered question

2021-02-22

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

Arnold Odonnell

Skilled2021-02-24Added 109 answers

Step 1
Given equation is

x + y^{2} = 1

.
To find second derivatives

\frac{d^{2} y}{{d x}^{2}}

.
Solution:
Power rule of differentiation states that:

\frac{d}{d x} (x^{n}) = n x^{n - 1}

Chain rule of differentiation states that:

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

Step 2
Differentiating the given function with respect to x,

\frac{d}{d x} (x + y^{2}) = \frac{d}{d x} (1)

1 + 2 y \frac{d y}{d x} = 0

2 y \frac{d y}{d x} = - 1

\frac{d y}{d x} = - \frac{1}{2 y}

Again differentiating with respect to x,

\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} (- \frac{1}{2 y})

= - \frac{1}{2} \frac{d}{d x} (\frac{1}{y})

= - \frac{1}{2} \cdot \frac{- 1}{y^{2}} \frac{d y}{d x}

= \frac{1}{2 y^{2}} \cdot \frac{1}{- 2 y}

= - \frac{1}{4 y^{3}} [U \sin g \frac{d y}{d x} = - \frac{1}{2 y}]

Step 3
Hence,

\frac{d^{2} y}{{d x}^{2}} = - \frac{1}{4 y^{3}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?