Question

Find y'' by implicit differentiation. x2 + 5y2 = 5

Answer 1

The expression for $y''$ is $\boxed{y''=\dfrac{-1}{5y^{3}}}$.

Further explanation:

Given:

The expression is $x^{2}+5y^{2}=5$.

Concept used:

Implicit Differentiation is used to differentiate the function which contains $y$ as a function of $x$ or $x$ is a function of $y$.

The implicit differentiation is useful to obtain the expression of $y'$ and $y''$.

The chain rule is the best tool in the implicit differentiation of the any expression.

The quotient rule of differentiation  is as follows:

$\boxed{\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v(du)-u(dv)}{v^{2}}}$

Calculation:

Differentiate the equation $x^{2}+5y^{2}=5$ with respect to $x$.

$\begin{aligned}\dfrac{d}{dx}(x^{2}+5y^{2})&=\dfrac{d}{dx}5\\2x+10y \dfrac{dy}{dx}&=0\\2x+10yy'&=0\end{aligned}$

Arrange the above equation to obtain the expression of $y'$.

$y'=-\dfrac{2x}{10y}$

Simplified the above equation to obtain the value of $y'$.

$y'=-\dfrac{x}{5y}$

Again differentiate the above equation with respect to the $x$ as follows:

$y''=-\dfrac{d}{dx}\left(\dfrac{x}{5y}\right)$

Apply the quotient rule of differentiation to obtain the value of $y''$.

$\begin{aligned}y'&'=-\dfrac{5y(1)-5xy'}{(5y)^{2}}\\&=\dfrac{-y+xy'}{5y^{2}}\end{aligned}$

Substitute $y'=-\frac{x}{5y}$ in the above equation.

$\begin{aligned}y''&=\dfrac{-y+xy'}{5y^{2}}\\&=\dfrac{-y+x(-\frac{x}{5y})}{5y^{2}}\\&=\dfrac{-5y^{2}-x^{2}}{25y^{3}}\\&=\dfrac{-(x^{2}+5y^{2})}{25y^{3}}\end{aligned}$

Substitute $x^{2}+5y^{2}=5$ in the above equation.

$\begin{aligned}y''&=\dfrac{-(x^{2}+5y^{2})}{25y^{3}}\\&=\dfrac{-5}{25y^{3}}\\&=\dfrac{-1}{5y^{3}}\end{aligned}$

Therefore, the expression of $y''$ by the implicit differentiation is $\boxed{y''=\dfrac{-1}{5y^{3}}}$.

Learn more:

1. Function: brainly.com/question/2142762

2. Quadratic equation: brainly.com/question/1332667

Answer details:

Grade: Senior school.

Subject: Mathematics.

Chapter: Differentiation.

Keywords: Implicit differentiation, explicit differentiation, chain rule, quotient rule of differentiation, the value of y', the value of y'', with respect to x,  chain rule.

Answer 2

Differentiate with respect to x and y
2x +10 y y' = 0
get y' alone
y' = -2x/(10 y)
reduce
y'=-x/(5y)
continue with quotient rule of differentiation
y'' = ( 5y(-1) - (-x) (5y')) / (5y) squared

y'' = (-5y + 5xy') /  (25y squared)
reduce by dividing each term by 5
y" = (-y +xy') / (5y squared)   answer

Now if the teacher wants you to continue so no y or y' shows
substitute  y' = -x/(5y) and y = square root ((5-x squared) / 5)
and simplify