How do you find $(dy)/(dx)$ given $x^3+y+8x=2y^2$ ?

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Answer 1 · 2017-06-01T20:53:39

$dy/dx = (3x^2+8)/(4y-1)$

Use implicit differentiation and the shorthand $dy/dx = y'$

$x^3+y+8x=2y^2$

$d/dx(x^3+y+8x) = d/dx(2y^2)$

$3x^2+y'+8 = 4y*y'$

Now use algebra to solve for $y'$ :

$3x^2+8 = 4y*y'-y'$

$3x^2+8 = y'(4y-1)$

$(3x^2+8)/(4y-1) = y'$

Therefore:

$dy/dx = (3x^2+8)/(4y-1)$

Answer 2 · 2017-06-01T20:54:31

$dy/dx = (3x^2+8)/(4y-1)$

Write the equation as:

$x^3+8x = 2y^2-y$

Differentiate both sides with respect to $x$ :

$d/dx (x^3+8x) = d/dx (2y^2-y)$

$3x^2+8 = (4y-1)dy/dx$

So that:

$dy/dx = (3x^2+8)/(4y-1)$

How do you find (dy)/(dx)(dy)/(dx) given x^3+y+8x=2y^2x^3+y+8x=2y^2?