Question

`x^(2) + y^(2) = 1\\`


find
`(d^(2) y)/(d x^(2))`

Answer 1

`\frac{\text{d}^2y}{\text{d}x^2}=-(x^2+y^2)/(y^3)`

Explanation:

Given equation:

`x^2+y^2=1`

To differentiate an equation that contains a mixture of x and y terms, we can use implicit differentiation.

Begin by placing d/dx in front of each term of the equation:

`\frac{\text{d}}{\text{d}x}x^2+\frac{\text{d}}{\text{d}x}y^2=\frac{\text{d}}{\text{d}x}1`

Differentiate the terms in x only (and constant terms) using the power rule and the constant rule:

`2x+\frac{\text{d}}{\text{d}x}y^2=0`

Use the chain rule to differentiate terms in y only.

In practice, this means differentiate with respect to y, and place dy/dx at the end:

`2x+2y\frac{\text{d}y}{\text{d}x}=0`

Rearrange to make dy/dx the subject:

`2y\frac{\text{d}y}{\text{d}x}=-2x`

`\frac{\text{d}y}{\text{d}x}=(-2x)/(2y)`

`\frac{\text{d}y}{\text{d}x}=-(x)/(y)`

To find the second derivative, differentiate again using the quotient rule.

`\boxed{\begin{array}{c}\underline{\textsf{Quotient\;Rule\;for\;Differentiation}}\\\\\textsf{If\;\;$y=(u)/(v)$\;\;then:}\\\\\frac{\text{d}y}{\text{d}x}=\frac{v \frac{\text{d}u}{\text{d}x}-u\frac{\text{d}v}{\text{d}x}}{v^2}\\\\\end{array}}`

`\textsf{Let\;$u=-x$}\implies \frac{\text{d}u}{\text{d}x}=-1`

`\textsf{Let\;$v=y$}\implies \frac{\text{d}v}{\text{d}x}=1\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}x}`

Put everything into the formula:

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

`\frac{\text{d}^2y}{\text{d}x^2}=\frac{-y+x\frac{\text{d}y}{\text{d}x}}{y^2}`

Substitute in the previously found dy/dx = -x/y:

`\frac{\text{d}^2y}{\text{d}x^2}=(-y+x\left(-(x)/(y)\right))/(y^2)`

Simplify:

`\frac{\text{d}^2y}{\text{d}x^2}=(-y-(x^2)/(y))/(y^2)`

`\frac{\text{d}^2y}{\text{d}x^2}=((-y^2)/(y)-(x^2)/(y))/(y^2)`

`\frac{\text{d}^2y}{\text{d}x^2}=((-y^2-x^2)/(y))/(y^2)`

`\frac{\text{d}^2y}{\text{d}x^2}=(-y^2-x^2)/(y^3)`

`\frac{\text{d}^2y}{\text{d}x^2}=-(x^2+y^2)/(y^3)`

Therefore, the second derivative of x² + y² = 1 is:

`\large\boxed{\boxed{\frac{\text{d}^2y}{\text{d}x^2}=-(x^2+y^2)/(y^3)}}`

Answer 2

`\sf (d^2y)/(dx^2) = \boxed{ (-(x^2+y^2))/(y^3)}`

Explanation:

To find the second derivative x² + y² = 1, we can use implicit differentiation.

Step 1: Find the first derivative
`\sf (dy)/(dx)` using implicit differentiation.

`\begin{aligned} x^2 + y^2 &= 1 \\ (d)/(dx)(x^2 + y^2) &= (d)/(dx)(1) \\ 2x + 2y(dy)/(dx) &= 0 \\ (dy)/(dx) &= -(x)/(y) \end{aligned}`

Step 2: Find the second derivative
`\sf (d^2y)/(dx^2)` by differentiating
`\sf (dy)/(dx)` with respect to x.

`\begin{aligned} (d)/(dx) \left( (dy)/(dx) \right) &= (d)/(dx) \left( -(x)/(y) \right) \\ (d^2y)/(dx^2) &= (-1 \cdot y - (-x)(dy)/(dx))/(y^2) \\ (d^2y)/(dx^2) &= (-y + x\cdot -\left((x)/(y)\right))/(y^2) \\ (d^2y)/(dx^2) &=( (-y\cdot y - x\cdot x )/(y))/(y^2) \\ (d^2y)/(dx^2) &= (- y^2-x^2)/(y^3) \\ (d^2y)/(dx^2) &= (- (x^2+y^2))/(y^3) \end{aligned}`

Therefore,
`\sf (d^2y)/(dx^2) = \boxed{ (- (x^2+y^2))/(y^3)}` is the second derivative of y with respect to x in terms of y.

Note that this formula is only valid for points where
`\sf y \\eq 0`.

Definition:

Implicit differentiation is a method of differentiating an equation where the function y is not explicitly defined in terms of x.

To differentiate implicitly, we take the derivative of both sides of the equation according to x, and treat y as an implicit function of x.