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

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

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

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

3 x^{2} - 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

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

Simplify

= 3 x^{2} - 2 x

\int x e^{x^{2} - 1} d x

\int_{0}^{1} e^{x^{4}} d x

\int cos^{4} (x t a) n^{3} x d x

e^{- h}

(- 2, - 7), (2, - 1)

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