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

The answer to (\partial)/(\partial x)(x^2-2xy+3y^2x^3) is 2x-2y+9y^2x^2

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

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

2 x - 2 y + 9 y^{2} x^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (3 y^{2} x^{3}) = 9 y^{2} x^{2}

= 2 x - 2 y + 9 y^{2} x^{2}

y = ln (\frac{x ^{2} - 1}{x ^{2} + 1})

\frac{d y}{d x} + 12 y = 6

- 10 e^{- 5 x^{2}} x

\int 6 x ln (5 x) d x

y = x^{4} - 4 x^{2} + 4, y = x^{2}

What is (\partial)/(\partial x)(x^2-2xy+3y^2x^3) ?
The answer to (\partial)/(\partial x)(x^2-2xy+3y^2x^3) is 2x-2y+9y^2x^2