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

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

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

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

6 y - 4 x

Solution steps

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

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

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

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

= 0 - 4 x + 6 y

Simplify

= 6 y - 4 x

\frac{d}{d x} ((4 x)^{x})

\int 10 x^{4} - 2 sec^{2} (x) d x

\frac{4}{x ^{2}}

\frac{9}{x}

x^{2} + x y - \frac{y ^{2}}{4} = 1

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