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

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

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

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

- x^{2} - x + 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (x^{3}) = 0

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

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

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

= 0 + 2 y - x^{2} - x

Simplify

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

x \to 1 - lim ((x - 1) ∣ x - 1 ∣)

f (x) = 4 - 6 x

\int_{2}^{2} x^{3} d x

\frac{d y}{d t} = \frac{t e ^{t}}{y 1 + y ^{2}}

\frac{d}{d x} (- 16 a x^{2} + a + \frac{1}{3})

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