What is (\partial)/(\partial y)((z)(6xy+y^2+x^2)) ?

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

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(\partial)/(\partial y)((z)(6xy+y^2+x^2))

\frac{\partial}{\partial y} ((z) (6 x y + y^{2} + x^{2}))

z (6 x + 2 y)

Solution steps

\frac{\partial}{\partial y} ((z) (6 x y + y^{2} + x^{2}))

Treat

z, x

as constants

Take the constant out:

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

= z \frac{\partial}{\partial y} (6 x y + y^{2} + x^{2})

Apply the Sum/Difference Rule:

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

= z (\frac{\partial}{\partial y} (6 x y) + \frac{\partial}{\partial y} (y^{2}) + \frac{\partial}{\partial y} (x^{2}))

\frac{\partial}{\partial y} (6 x y) = 6 x

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

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

= z (6 x + 2 y + 0)

Simplify

= z (6 x + 2 y)

\frac{x ^{2} ( 4 - x ^{2} )}{2}

50 \cdot (cos (\frac{π}{3}))^{'}

\frac{1}{( x ^{2} + 2 ) ^{3}}

\frac{x ( 1000 - 2 x )}{5}

\frac{2 - 6 x - 3 x ^{2}}{( 1 + x ) ( 1 - 2 x ) ^{2}}

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