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

The answer to (\partial)/(\partial x)(x^2y^2-x) is 2y^2x-1

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

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

2 y^{2} x - 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y^{2} x - 1

\int sin^{2} (4 x) \cdot cos^{3} (4 x) d x

\frac{d}{d x} (ln^{2} (3) \cdot 3^{x})

x \to 2 lim (\frac{3 x - 2}{1 - 2 x})

x \frac{d v}{d x} = 2 v + \frac{10}{v}

x \to 7 lim (x^{2} - 4 x + 18)

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