if [tex]x^2+y^2=25[/tex], what is the value of [tex]\frac{d^2y}{dx^2}[/tex] at the point (4,3)?(adsbygoogle = window.adsbygoogle || []).push({});

[tex]x^2+y^2=25[/tex]

[tex]2x+2y\frac{dy}{dx}=0[/tex]

[tex]\frac{dy}{dx}=-\frac{2x}{2y}[/tex]

[tex]\frac{dy}{dx}=-\frac{3}{4}[/tex]

[tex]\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(-x)}{x^2}[/tex]

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

[tex]\frac{d^2y}{dx^2}=0[/tex]

where did I go wrong?

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# Homework Help: Dif. equ.

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