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

The answer to (\partial)/(\partial x)(3x^2y-y^3+x-y) is 6xy+1

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

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

6 x y + 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (x) = 1

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

= 6 y x - 0 + 1 - 0

Simplify

= 6 x y + 1

x \to 0 lim (x^{3} cos (\frac{2}{x}))

\int \frac{30}{( x - 1 ) ( x ^{2} + 9 )} d x

f (x) = \frac{4 x ^{3}}{x ^{4} + 1}

\frac{d}{d z} (ln (x^{2} + y^{2} + z^{2}))

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

What is (\partial)/(\partial x)(3x^2y-y^3+x-y) ?
The answer to (\partial)/(\partial x)(3x^2y-y^3+x-y) is 6xy+1