What is derivative of 2y^3x+x^3=y^5+12 ?

The answer to derivative of 2y^3x+x^3=y^5+12 is (dy)/(dx)=(-2y^3-3x^2)/(y^2(-5y^2+6x))

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derivative of 2y^3x+x^3=y^5+12

derivative of

2 y^{3} x + x^{3} = y^{5} + 12

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

Solution steps

2 y^{3} x + x^{3} = y^{5} + 12

Treat

y

y (x)

Differentiate both sides:

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

Isolate

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

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

What is derivative of 2y^3x+x^3=y^5+12 ?
The answer to derivative of 2y^3x+x^3=y^5+12 is (dy)/(dx)=(-2y^3-3x^2)/(y^2(-5y^2+6x))