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}

x^{3} + 4 x

y = 7 cos (2 x), y = 7 - 7 cos (2 x)

\frac{d}{d x} (5 x^{2})

\frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

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

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