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

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

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

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

3 x^{2} - 2 y

Solution steps

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

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

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

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

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

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

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

Simplify

= 3 x^{2} - 2 y

y^{^{'''}} = 0

\frac{4}{5 s}

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

x sin (x)

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