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

The answer to (\partial)/(\partial x)((y)(x^3-2y^2)) is 3yx^2

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(\partial)/(\partial x)((y)(x^3-2y^2))

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

3 y x^{2}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= 3 y x^{2}

5 a + \frac{5}{4} - 3 - 7 \frac{a}{6}

\frac{3 ( x + 5 )}{x + 4}

\frac{25}{s ( s + 5 )}

\frac{- ( X - 10 ) ^{3}}{375000}

so l v e (- 1575000 + \frac{x}{1 + ( 0.2139600000000 \dots ) ^{2}} = 0, x)

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