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

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

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

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

6 x y^{4}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{2} y^{4}) = 6 y^{4} x

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

= 6 y^{4} x + 0

Simplify

= 6 x y^{4}

f (x) = \frac{cos ( 2 x ) ln ( x )}{x - 5}

\int x \cdot e^{6 x} d x

2 (x - 1)^{2} + 50, 3 (x + 3)^{2} + 2

\int x^{2} - 5 x d x

\int_{0}^{\infty} 4 sin^{2} (x) d x

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