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

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

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

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

2 y - 3 x

Solution steps

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

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

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

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

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

= 0 - 3 x + 2 y

Simplify

= 2 y - 3 x

f (x) = 2 x^{2} - x

\int \frac{1}{t ^{2} - 2 t + 26} d t

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

\int_{\frac{1}{2}}^{5} 12 x ln (2 x) d x

\frac{3}{x}

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