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

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05-07-2017
Mathematics

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

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marycaresmuch marycaresmuch · Answer 1 · 2017-07-05T13:06:58+02:00

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

AkhileshT AkhileshT · Answer 2 · 2019-09-06T09:52:22+02:00

The expression for [tex]y''[/tex] is [tex]\boxed{y''=\dfrac{-1}{5y^{3}}}[/tex].

Further explanation:

Given:

The expression is [tex]x^{2}+5y^{2}=5[/tex].

Concept used:

Implicit Differentiation is used to differentiate the function which contains [tex]y[/tex] as a function of [tex]x[/tex] or [tex]x[/tex] is a function of [tex]y[/tex].

The implicit differentiation is useful to obtain the expression of [tex]y'[/tex] and [tex]y''[/tex].

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

The quotient rule of differentiation is as follows:

[tex]\boxed{\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v(du)-u(dv)}{v^{2}}}[/tex]

Calculation:

Differentiate the equation [tex]x^{2}+5y^{2}=5[/tex] with respect to [tex]x[/tex].

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

Arrange the above equation to obtain the expression of [tex]y'[/tex].

[tex]y'=-\dfrac{2x}{10y}[/tex]

Simplified the above equation to obtain the value of [tex]y'[/tex].

[tex]y'=-\dfrac{x}{5y}[/tex]

Again differentiate the above equation with respect to the [tex]x[/tex] as follows:

[tex]y''=-\dfrac{d}{dx}\left(\dfrac{x}{5y}\right)[/tex]

Apply the quotient rule of differentiation to obtain the value of [tex]y''[/tex].

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

Substitute [tex]y'=-\frac{x}{5y}[/tex] in the above equation.

[tex]\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}[/tex]

Substitute [tex]x^{2}+5y^{2}=5[/tex] in the above equation.

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

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

Learn more:

1. Function: https://brainly.com/question/2142762

2. Quadratic equation: https://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.