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

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

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

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

6 x y^{4}

Solution steps

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

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} (2)

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

\frac{\partial}{\partial x} (2) = 0

= 6 y^{4} x - 0

Simplify

= 6 x y^{4}

\int (5 x - 7) e^{5 x - 7} d x

\int t e^{- 3 t} d t

g (t) = \frac{t - t}{t ^{\frac{1}{9}}}

x^{2} y^{^{''}} + 7 x y^{^{'}} + 5 y = 0

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

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