What is (\partial)/(\partial x)(4y^2x^3-3yx^2) ?

The answer to (\partial)/(\partial x)(4y^2x^3-3yx^2) is 12y^2x^2-6yx

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(\partial)/(\partial x)(4y^2x^3-3yx^2)

\frac{\partial}{\partial x} (4 y^{2} x^{3} - 3 y x^{2})

12 y^{2} x^{2} - 6 y x

Solution steps

\frac{\partial}{\partial x} (4 y^{2} x^{3} - 3 y x^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4 y^{2} x^{3}) - \frac{\partial}{\partial x} (3 y x^{2})

\frac{\partial}{\partial x} (4 y^{2} x^{3}) = 12 y^{2} x^{2}

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

= 12 y^{2} x^{2} - 6 y x

x^{^{''}} + 4 x^{^{'}} + 2 x = 0

\int (5 e^{- 0.8 x}) d x

y = 2 x e^{2 x} - e^{2 x}

\int 18 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

(x - 8 y) d x + (4 x) d y = 0

What is (\partial)/(\partial x)(4y^2x^3-3yx^2) ?
The answer to (\partial)/(\partial x)(4y^2x^3-3yx^2) is 12y^2x^2-6yx