What is (\partial)/(\partial x)(-x^3+4xy-2y^2+1) ?

The answer to (\partial)/(\partial x)(-x^3+4xy-2y^2+1) is 4y-3x^2

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(\partial)/(\partial x)(-x^3+4xy-2y^2+1)

\frac{\partial}{\partial x} (- x^{3} + 4 x y - 2 y^{2} + 1)

4 y - 3 x^{2}

Solution steps

\frac{\partial}{\partial x} (- x^{3} + 4 x y - 2 y^{2} + 1)

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} (4 x y) - \frac{\partial}{\partial x} (2 y^{2}) + \frac{\partial}{\partial x} (1)

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

\frac{\partial}{\partial x} (4 x y) = 4 y

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

\frac{\partial}{\partial x} (1) = 0

= - 3 x^{2} + 4 y - 0 + 0

Simplify

= 4 y - 3 x^{2}

x \to 7 lim (2 x + 5)

\frac{d}{d x} ((x + 4)^{2})

\int (5 x - 1)^{7} d x

2 x, 9 x - x^{2}, (7, 14)

f (x) = \frac{6}{ln ( x ^{2} + 4 )}

What is (\partial)/(\partial x)(-x^3+4xy-2y^2+1) ?
The answer to (\partial)/(\partial x)(-x^3+4xy-2y^2+1) is 4y-3x^2