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

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

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

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

2 y x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (y^{3}) = 0

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

\frac{\partial}{\partial x} (3 y) = 0

= 0 + 2 y x - 0

Simplify

= 2 y x

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

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

f (x) = x - 2, at x = 3

cos (2 x) + sin (x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 0

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