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

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

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

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

2 x y + 2 x y^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 y x - 0 + 2 y^{3} x

Simplify

= 2 x y + 2 x y^{3}

\int 1 - 4 x d x

cot^{2} (sin (θ))

y = cos (sin (tan (π x)))

\frac{d y}{d x} = y^{(\frac{1}{3})}

\int_{0}^{7} 7 - x d x

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