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

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

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

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

2 x y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y^{2} x - 0

Simplify

= 2 x y^{2}

\int 3 x + 2 d x

f (x) = 4 x + 8, 0, 8

n = 0 \sum \infty \frac{7 n !}{2 ^{n}}

f (x) = 1 - x - x^{2}, at x = 0

x (29 - 2 x) (16 - 2 x)

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