What is implicit (dy)/(dx),-x+y-x^3=y^3+4x^2 ?

The answer to implicit (dy)/(dx),-x+y-x^3=y^3+4x^2 is (dy)/(dx)=(8x+3x^2+1)/(1-3y^2)

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implicit (dy)/(dx),-x+y-x^3=y^3+4x^2

implicit derivative

\frac{d y}{d x}, - x + y - x^{3} = y^{3} + 4 x^{2}

\frac{d y}{d x} = \frac{8 x + 3 x ^{2} + 1}{1 - 3 y ^{2}}

Solution steps

- x + y - x^{3} = y^{3} + 4 x^{2}

Treat

y

y (x)

Differentiate both sides:

- 1 + \frac{d y}{d x} - 3 x^{2} = 3 y^{2} \frac{d y}{d x} + 8 x

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{8 x + 3 x ^{2} + 1}{1 - 3 y ^{2}}

\frac{d y}{d x} = \frac{8 x + 3 x ^{2} + 1}{1 - 3 y ^{2}}

\int \frac{( x + 1 ) ^{2}}{x ^{2}} d x

\int \frac{1}{n ( n + 1 )}

y^{^{''}} - 3 y^{^{'}} - 18 y = 0

x^{3} - 15 x^{2} + 50 x, - x^{3} + 15 x^{2} - 50 x

n = 0 \sum \infty \frac{1}{2 n + 3}

What is implicit (dy)/(dx),-x+y-x^3=y^3+4x^2 ?
The answer to implicit (dy)/(dx),-x+y-x^3=y^3+4x^2 is (dy)/(dx)=(8x+3x^2+1)/(1-3y^2)