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

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

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

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

- 4 x - 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

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

= 0 - 4 x - 1 - 0 + 0

Simplify

= - 4 x - 1

\int_{0}^{1} \frac{x}{1 + x} d x

\frac{d}{d t} (e^{3 t})

\frac{d}{d x} (a cos (t) + b sin (t))

y = 6 x, y = \frac{2}{3} x, y = 40 - x^{2}

\int x (2 x^{2} + 1)^{\frac{3}{2}} d x

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