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

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

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

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

- 2 x - 1

Solution steps

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

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

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

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

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

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

= 0 - 2 x - 1 - 0 + 0

Simplify

= - 2 x - 1

ln (2) + ln (x)

y = \frac{e ^{6 x}}{e ^{6 x} + 3}

\frac{\partial}{\partial x} (e^{- 3 y} cos (π x))

6 y^{^{''}} - y^{^{'}} - 2 y = 0

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

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