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

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

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

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

x^{2} + 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (y^{2}) = 2 y

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

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

= 0 + 2 y + x^{2} + 0

Simplify

= x^{2} + 2 y

\int \frac{x ^{2}}{x ^{6} + 1} d x

f (x) = x^{3} - x

\frac{d}{d x} ((x^{2} - 8) (5 x^{3}))

\int \frac{cos ( ln ( 2 x ) )}{x} d x

\int \frac{x ^{2} + 16 x - 4}{x ^{3} - 4 x} d x

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