What is (\partial)/(\partial x)(6y(3x^2-y^2)) ?

The answer to (\partial)/(\partial x)(6y(3x^2-y^2)) is 36yx

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(\partial)/(\partial x)(6y(3x^2-y^2))

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

36 y x

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

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

= 6 y (6 x - 0)

Simplify

= 36 y x

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

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

\frac{d}{d x} ((5 + x^{4})^{12})

\int \frac{1}{v ^{2} + 1} d v

\frac{2 s + 1}{3 s ^{2} + 5 s}

What is (\partial)/(\partial x)(6y(3x^2-y^2)) ?
The answer to (\partial)/(\partial x)(6y(3x^2-y^2)) is 36yx