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

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

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

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

3 y^{2} + 4 - 4 x

Solution steps

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

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

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

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

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

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

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

Simplify

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

\int (2 x yz + sin (x)) d x

\frac{d}{d x} (x^{4} (x^{2} + 1)^{6})

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

\frac{1}{( 7 + x ) 2}

x \to 0 lim (x sin (x))

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