What is (\partial)/(\partial x)(9-2x+4y-x^2-4y^2) ?

The answer to (\partial)/(\partial x)(9-2x+4y-x^2-4y^2) is -2x-2

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(\partial)/(\partial x)(9-2x+4y-x^2-4y^2)

\frac{\partial}{\partial x} (9 - 2 x + 4 y - x^{2} - 4 y^{2})

- 2 x - 2

Solution steps

\frac{\partial}{\partial x} (9 - 2 x + 4 y - x^{2} - 4 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (9) - \frac{\partial}{\partial x} (2 x) + \frac{\partial}{\partial x} (4 y) - \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (4 y^{2})

\frac{\partial}{\partial x} (9) = 0

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

\frac{\partial}{\partial x} (4 y) = 0

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

\frac{\partial}{\partial x} (4 y^{2}) = 0

= 0 - 2 + 0 - 2 x - 0

Simplify

= - 2 x - 2

f (x) = \frac{3}{x} + x, 5, 15

\frac{d}{d θ} (cos (θ) sin (θ))

\frac{1}{2 t - 1}

\int (1 - t) (7 + t^{2}) d t

x \to 1 + lim (2 - x)

What is (\partial)/(\partial x)(9-2x+4y-x^2-4y^2) ?
The answer to (\partial)/(\partial x)(9-2x+4y-x^2-4y^2) is -2x-2