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

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

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

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

y (3 x^{2} - 6 y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= y (3 x^{2} - 6 y)

\frac{4 z ^{2}}{( z - 1 ) ^{2}}

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

6 (3 x [- 5])

\frac{e ^{- 3 s} + 2 s}{s ( s + 1 )}

- \frac{3 j x ( a e ^{\frac{π}{t}} - b e ^{\frac{π}{t}} )}{2}

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