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

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

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

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

6 x y^{4}

Solution steps

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

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

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

= 6 y^{4} x - 0

Simplify

= 6 x y^{4}

\int \frac{1}{sin ( x ) + cos ( x )} d x

\int_{- 1}^{1} \frac{1}{2 x + 3} d x

\frac{d}{d x} (\frac{1}{3} x^{(\frac{1}{2})} - x^{(\frac{3}{2})})

f (x) = \frac{1}{x}, at x = \frac{1}{4}

\int (ln (t)) d t

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