What is (\partial)/(\partial x)(3x^2-2xy+y^2-8y) ?

The answer to (\partial)/(\partial x)(3x^2-2xy+y^2-8y) is 6x-2y

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(\partial)/(\partial x)(3x^2-2xy+y^2-8y)

\frac{\partial}{\partial x} (3 x^{2} - 2 x y + y^{2} - 8 y)

6 x - 2 y

Solution steps

\frac{\partial}{\partial x} (3 x^{2} - 2 x y + y^{2} - 8 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (2 x y) + \frac{\partial}{\partial x} (y^{2}) - \frac{\partial}{\partial x} (8 y)

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

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

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

\frac{\partial}{\partial x} (8 y) = 0

= 6 x - 2 y + 0 - 0

Simplify

= 6 x - 2 y

x \to \infty lim (x + x - x - x)

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

F (x) = 3 x^{4} - 65 x^{3}

t + 1

\int \frac{( 1 + cos ( 2 x ) )}{2} d x

What is (\partial)/(\partial x)(3x^2-2xy+y^2-8y) ?
The answer to (\partial)/(\partial x)(3x^2-2xy+y^2-8y) is 6x-2y