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

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

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

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

2 y - 4 + x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 y - 4 + x

\int (4 x^{2} - 9 x)^{5} (8 x - 9) d x

x \to 0 lim (\frac{sin ^{3} ( x )}{x ^{2}})

n = 1 \sum \infty \frac{n}{e ^{4 n}}

\int \frac{2 x ^{2} - 5 x + 2}{x ^{3} + x} d x

\frac{d}{d x} (\frac{- 2}{( x - 1 ) ^{3}})

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