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

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

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

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

6 x y^{2}

Solution steps

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

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 y)

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

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

= 6 y^{2} x - 0

Simplify

= 6 x y^{2}

x \to \infty lim ((1 + \frac{2}{x + 1})^{x + 1})

n = 0 \sum \infty n^{- 0.48}

x \to 1 lim ([f (x) + g (x)])

\frac{d}{d x} ((10 x)^{l n (10 x)})

\frac{\partial}{\partial x} (tan (x - y))

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