How do you implicitly differentiate $x^{3} - 3 x y + 2 y^{3} = 3$ $x^3 - 3xy + 2y^3 = 3$ ?

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How do you implicitly differentiate $x^{3} - 3 x y + 2 y^{3} = 3$ $x^3 - 3xy + 2y^3 = 3$ ?

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Answer 1 · 2015-12-05T15:52:13

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

Explanation:

We assume that y is a function of x, ie $y = f (x)$ $y=f(x)$ and then differentiate each side of the equation with respect to x, then re-arrange and solve for $\frac{d y}{d x}$ $dy/dx$ .

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

$therefore 3x^2-3xdy/dx-3y+6y^2dy/dx=0$ . (Used product rule and power rule)

$therefore dy/dx(-3x+6y^2)=3y-3x^2$ . (Took out common factor)

$therefore dy/dx=(3y-3x^2)/(6y^2-3x)$

$therefore dy/dx=(y-x^2)/(2y^2-x)$ (Divide all by 3).

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How do you implicitly differentiate $x^{3} - 3 x y + 2 y^{3} = 3$ $x^3 - 3xy + 2y^3 = 3$ ?

1 Answer

Explanation:

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Science

Math

Humanities

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How do you implicitly differentiate x^3 - 3xy + 2y^3 = 3x3−3xy+2y3=3 x^3 - 3xy + 2y^3 = 3?

1 Answer

Explanation:

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How do you implicitly differentiate $x^{3} - 3 x y + 2 y^{3} = 3$ $x^3 - 3xy + 2y^3 = 3$ ?