What is (\partial)/(\partial x)((x^2+2y)/(x^3-y^2)) ?

The answer to (\partial)/(\partial x)((x^2+2y)/(x^3-y^2)) is (-x^4-6x^2y-2xy^2)/((x^3-y^2)^2)

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(\partial)/(\partial x)((x^2+2y)/(x^3-y^2))

\frac{\partial}{\partial x} (\frac{x ^{2} + 2 y}{x ^{3} - y ^{2}})

\frac{- x ^{4} - 6 x ^{2} y - 2 x y ^{2}}{( x ^{3} - y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x ^{2} + 2 y}{x ^{3} - y ^{2}})

Treat

y

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{\partial}{\partial x} ( x ^{2} + 2 y ) ( x ^{3} - y ^{2} ) - \frac{\partial}{\partial x} ( x ^{3} - y ^{2} ) ( x ^{2} + 2 y )}{( x ^{3} - y ^{2} ) ^{2}}

\frac{\partial}{\partial x} (x^{2} + 2 y) = 2 x

\frac{\partial}{\partial x} (x^{3} - y^{2}) = 3 x^{2}

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

Expand

2 x (x^{3} - y^{2}) - 3 x^{2} (x^{2} + 2 y) : - x^{4} - 6 x^{2} y - 2 x y^{2}

= \frac{- x ^{4} - 6 x ^{2} y - 2 x y ^{2}}{( x ^{3} - y ^{2} ) ^{2}}

\frac{d}{d x} (ln (x x^{2} - 5))

n = 0 \sum \infty \frac{x ^{n}}{4 ^{n}}

2 x^{2}, at 2

\frac{d}{d x} (x e^{2 x - y^{2}})

\frac{d y}{d x} - y = x

What is (\partial)/(\partial x)((x^2+2y)/(x^3-y^2)) ?
The answer to (\partial)/(\partial x)((x^2+2y)/(x^3-y^2)) is (-x^4-6x^2y-2xy^2)/((x^3-y^2)^2)