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

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

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

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

2 x y^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y^{3} x - 0

Simplify

= 2 x y^{3}

\frac{d y}{d x}, 4 x y = 8

\int \frac{2}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

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

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

\int + \frac{4 - 32 x}{1 - 16 x ^{2}} d x

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