What is the solution for derivative of 4x^3+2y^5=6 ?

The solution for derivative of 4x^3+2y^5=6 is y=\sqrt[5]{3x+c_{1}-2x^3}

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derivative of 4x^3+2y^5=6

\frac{d}{d x} (4 x^{3} + 2 y^{5}) = 6

y = 5 3 x + c_{1} - 2 x^{3}

Solution steps

\frac{d}{d x} (4 x^{3} + 2 y^{5}) = 6

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

4 x^{3} + 2 y^{5} = \int 6 d x

\int 6 d x = 6 x + c_{1}

4 x^{3} + 2 y^{5} = 6 x + c_{1}

Isolate

y : y = 5 3 x + \frac{c _{1}}{2} - 2 x^{3}

y = 5 3 x + \frac{c _{1}}{2} - 2 x^{3}

Simplify

y = 5 3 x + c_{1} - 2 x^{3}

\frac{d y}{d x} = (x - 1) (y + 1)

t^{2} y + 11 t y^{^{'}} - 11 y = 0, y^{1} (0) = t

\frac{d P}{d t} = 2 P - \frac{P ^{2}}{2}

\frac{d y}{d x} = (x - 1) (y + 2)

y^{^{'}} = \frac{x y}{1 + x ^{2}}

What is the solution for derivative of 4x^3+2y^5=6 ?
The solution for derivative of 4x^3+2y^5=6 is y=\sqrt[5]{3x+c_{1}-2x^3}