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

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

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(\partial)/(\partial x)(4yx^3-6y^2x^2)

\frac{\partial}{\partial x} (4 y x^{3} - 6 y^{2} x^{2})

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

Solution steps

\frac{\partial}{\partial x} (4 y x^{3} - 6 y^{2} x^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4 y x^{3}) - \frac{\partial}{\partial x} (6 y^{2} x^{2})

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

\frac{\partial}{\partial x} (6 y^{2} x^{2}) = 12 y^{2} x

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

f (x) = \frac{1}{1 + x ^{2}}, (1, \frac{1}{2})

\frac{d}{d y} (4)

\int 15 z^{3} e^{z} d z

e^{- 4 (t - 1)}

y^{^{'''}} + 4 y^{^{'}} = cot (2 x)

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