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

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

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

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

3 x^{2} - 6 x

Solution steps

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

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

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 3 x^{2} + 0 - 6 x + 0

Simplify

= 3 x^{2} - 6 x

\frac{d}{d a} (1 + cos^{4} (a) - sin^{4} (a))

x \to \infty lim (\frac{2 x}{x})

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{3 x} sin (2 x)

\frac{d y}{d x}, y = x^{3} + 2

x \to \infty lim (e^{0})

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