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

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

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

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

z (2 y - 3 x)

Solution steps

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

Treat

z, x

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

= z (\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

= z (0 - 3 x + 2 y)

Simplify

= z (2 y - 3 x)

\frac{d}{d x} ((8 x + 3) (x^{2} - 1)) + \frac{x + 4}{5}

(A x^{2} + B x + C) e^{3 x}

\frac{( 2 x ^{3} - 3 ) ^{6}}{36 x ^{2}}

f (t + 2 π)

\frac{7 \cdot 2 ^{k + 1} + ( - 1 ) ^{k + 2}}{3}

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