How do you find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ for $5 = 4 x^{3} - 4 y^{2}$ $5=4x^3-4y^2$ ?

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Answer 1 · 2016-11-04T17:47:06

$\frac{d^{2} y}{{d x}^{2}} = \frac{12 x y^{2} - 9 x^{4}}{4 y^{3}}$ $(d^2y)/(dx^2)=(12xy^2-9x^4)/(4y^3)$

$5 = 4 x^{3} - 4 y^{2}$ $5=4x^3-4y^2$

$0 = 12 x^{2} - 8 y \frac{d y}{d x}$ $0=12x^2-8y dy/dx$

$8 y \frac{d y}{d x} = 12 x^{2}$ $8ydy/dx=12x^2$

$\frac{d y}{d x} = \frac{12 x^{2}}{8 y} = \frac{3 x^{2}}{2 y}$ $dy/dx=(12x^2)/(8y)=(3x^2)/(2y)$

To find the second derivative use the quotient rule

$\frac{d^{2} y}{{d x}^{2}} = \frac{2 y \cdot 6 x - 3 x^{2} \cdot 2 \frac{d y}{d x}}{4 y^{2}}$ $(d^2y)/(dx^2)=(2y*6x-3x^2*2dy/dx)/(4y^2)$

$\frac{d^{2} y}{{d x}^{2}} = \frac{2 (6 x y - 3 x^{2} \frac{d y}{d x})}{4 y^{2}}$ $(d^2y)/(dx^2)=(2(6xy-3x^2dy/dx))/(4y^2)$

$\frac{d^{2} y}{{d x}^{2}} = \frac{6 x y - 3 x^{2} \cdot \frac{3 x^{2}}{2 y}}{2 y^{2}}$ $(d^2y)/(dx^2)=(6xy-3x^2*(3x^2)/(2y))/(2y^2)$

$\frac{d^{2} y}{{d x}^{2}} = \frac{6 x y - \frac{9 x^{4}}{2 y}}{2 y^{2}}$ $(d^2y)/(dx^2)=(6xy-(9x^4)/(2y))/(2y^2)$

$\frac{d^{2} y}{{d x}^{2}} = \frac{\frac{12 x y^{2} - 9 x^{4}}{2 y}}{2 y^{2}}$ $(d^2y)/(dx^2)=((12xy^2-9x^4)/(2y))/(2y^2)$

$\frac{d^{2} y}{{d x}^{2}} = (\frac{12 x y^{2} - 9 x^{4}}{2 y}) \cdot (\frac{1}{2 y^{2}})$ $(d^2y)/(dx^2)=((12xy^2-9x^4)/(2y))*(1/(2y^2))$

$\frac{d^{2} y}{{d x}^{2}} = \frac{12 x y^{2} - 9 x^{4}}{4 y^{3}}$ $(d^2y)/(dx^2)=(12xy^2-9x^4)/(4y^3)$

How do you find (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2) for 5=4x^3-4y^25=4x3−4y25=4x^3-4y^2?