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

The answer to (\partial)/(\partial y)(3x-2y^4) is -8y^3

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

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

- 8 y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (2 y^{4}) = 8 y^{3}

= 0 - 8 y^{3}

Simplify

= - 8 y^{3}

\int \frac{x ^{2}}{x ^{3} - 2} d x

x = - 2, x = 3, y = x^{2} - 12, y = x - 6

1 - x^{12}

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

\int x sin (\frac{1}{6} x) d x

What is (\partial)/(\partial y)(3x-2y^4) ?
The answer to (\partial)/(\partial y)(3x-2y^4) is -8y^3