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

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

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

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

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

Solution steps

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

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

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

\frac{\partial}{\partial x} (x - y) = 1

= \frac{3 x ^{2} ( x - y ) - 1 \cdot ( x ^{3} + y ^{2} )}{( x - y ) ^{2}}

Simplify

\frac{3 x ^{2} ( x - y ) - 1 \cdot ( x ^{3} + y ^{2} )}{( x - y ) ^{2}} : \frac{2 x ^{3} - 3 x ^{2} y - y ^{2}}{( x - y ) ^{2}}

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

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

\int_{1}^{2} (cos (2 x) - \frac{2}{x}) d x

f (x) = (- 4 x^{2} + 9)^{2}

\frac{d}{d x} (\frac{12}{x + 1})

x \to - 3 + lim (x)

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