What is (\partial)/(\partial x)(x^3+x^2y-y^2-4y) ?

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

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(\partial)/(\partial x)(x^3+x^2y-y^2-4y)

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

3 x^{2} + 2 x y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (4 y) = 0

= 3 x^{2} + 2 y x - 0 - 0

Simplify

= 3 x^{2} + 2 x y

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

x \to 5 + lim (\frac{2}{x + 5})

\frac{\partial}{\partial x} (a x + b y^{2} + c x^{2} y)

\frac{d}{d x} (arctan (tan (x)))

x \to 0 + lim (x ln (x^{2}))

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