What is (\partial)/(\partial y)(x^3-y^3+6xy) ?

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

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(\partial)/(\partial y)(x^3-y^3+6xy)

\frac{\partial}{\partial y} (x^{3} - y^{3} + 6 x y)

6 x - 3 y^{2}

Solution steps

\frac{\partial}{\partial y} (x^{3} - y^{3} + 6 x y)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{3}) - \frac{\partial}{\partial y} (y^{3}) + \frac{\partial}{\partial y} (6 x y)

\frac{\partial}{\partial y} (x^{3}) = 0

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

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

= 0 - 3 y^{2} + 6 x

Simplify

= 6 x - 3 y^{2}

\frac{d}{d x} (x y^{2} - x^{3} y) = 0

(- 21.2, 22.8), (- 20.2, 7.4)

n = 1 \sum \infty \frac{5}{π ^{n}}

\int \frac{( x + 4 )}{x ^{2} + 11 x + 18} d x

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

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