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

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

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

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

6 x^{3} y^{2} + 4

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (4 y) = 4

= 6 x^{3} y^{2} + 0 + 4

Simplify

= 6 x^{3} y^{2} + 4

\int_{- 1}^{1} e^{x} - (x^{2} - 1) d x

5 x^{2} + 2, 0, 2

f (t) = (t^{2} + 3 t + 7) (3 t^{2} + 3)

\int \frac{1}{x ^{2} 81 - x ^{2}} d x

\int x^{3} e^{3 x^{2}} d x

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