What is (\partial)/(\partial x)(3x^2y^4+1) ?

The answer to (\partial)/(\partial x)(3x^2y^4+1) is 6xy^4

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(\partial)/(\partial x)(3x^2y^4+1)

\frac{\partial}{\partial x} (3 x^{2} y^{4} + 1)

6 x y^{4}

Solution steps

\frac{\partial}{\partial x} (3 x^{2} y^{4} + 1)

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} (1)

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

\frac{\partial}{\partial x} (1) = 0

= 6 y^{4} x + 0

Simplify

= 6 x y^{4}

\int \frac{1}{x 9 x ^{2} - 4} d x

k = 1 \sum \infty k^{3} \cdot e^{- k^{2}}

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

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

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

What is (\partial)/(\partial x)(3x^2y^4+1) ?
The answer to (\partial)/(\partial x)(3x^2y^4+1) is 6xy^4