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

The answer to (\partial)/(\partial x)(3x^2y^2+4) is 6xy^2

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

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

6 x y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{2} y^{2}) = 6 y^{2} x

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

= 6 y^{2} x + 0

Simplify

= 6 x y^{2}

f (x) = \frac{2}{x - 4}, (6, 1)

\frac{d}{d x} ((4^{x^{3}} + 2)^{3})

\frac{\partial}{\partial x} (tan (\frac{x y}{z ^{2}}))

\int \frac{x ^{2} - x - 2}{x ^{3} - 2 x - 4} d x

4 cos^{4} (x)

What is (\partial)/(\partial x)(3x^2y^2+4) ?
The answer to (\partial)/(\partial x)(3x^2y^2+4) is 6xy^2