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)#