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Let y^4+5x=11y 4 +5x=11y, start superscript, 4, end superscript, plus, 5, x, equals, 11. What is the value of \dfrac{d^2y}{dx^2} dx 2 d 2 y ​ start fraction, d, squared, y, divided by, d, x, squared, end fraction at the point (2,1)(2,1)left parenthesis, 2, comma, 1, right parenthesis?

Answer :

abidemiokin

Answer:

[tex]\frac{d^2y}{dx^2} = \frac{-300y^2}{(4y^3-11)^3}[/tex]

Step-by-step explanation:

Given the expression  [tex]y^4+5x=11y[/tex], we are to find the second derivative [tex]\frac{d^2y}{dx^2}[/tex]

This differentiation will be implicit (indirect) as shown:

[tex]4y^3 \frac{dy}{dx} + 5 = 11\frac{dy}{dx} \\4y^3 \frac{dy}{dx} - 11 \frac{dy}{dx} = -5\\\frac{dy}{dx}(4y^3-11) = -5\\\frac{dy}{dx} = \frac{-5}{4y^3-11}[/tex]

Differentiating the second time using quotient rule:

[tex]\frac{d^2y}{dx^2} = \frac{4y^3-11 (0)- (-5)12y^2\frac{dy}{dx} }{(4y^3-11)^2} \\\\\frac{d^2y}{dx^2} = \frac{60y^2\frac{dy}{dx} }{(4y^3-11)^2} \\\\\frac{d^2y}{dx^2} = \frac{60y^2(\frac{-5}{4y^3-11} ) }{(4y^3-11)^2}\\\\\frac{d^2y}{dx^2} = \frac{-300y^2}{(4y^3-11)^3}[/tex]

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