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

The answer to (\partial)/(\partial y)(3x^2y^4-3) is 12x^2y^3

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

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

12 x^{2} y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (3 x^{2} y^{4}) = 12 x^{2} y^{3}

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

= 12 x^{2} y^{3} - 0

Simplify

= 12 x^{2} y^{3}

d er i v a t i v e f (x) = 4 x^{3}

\frac{d}{d x} (ln (1 + x y))

\int 30 (x + 7) (x + 5)^{4} d x

x \to 8 lim (\frac{8 x}{x - 8})

y^{^{'}} = (1 - 2 t) y^{2}, y (0) = - \frac{1}{6}

What is (\partial)/(\partial y)(3x^2y^4-3) ?
The answer to (\partial)/(\partial y)(3x^2y^4-3) is 12x^2y^3