What is (\partial)/(\partial x)(-400x(y-x^2)-2(1-x)) ?

The answer to (\partial)/(\partial x)(-400x(y-x^2)-2(1-x)) is -400(y-3x^2)+2

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(\partial)/(\partial x)(-400x(y-x^2)-2(1-x))

\frac{\partial}{\partial x} (- 400 x (y - x^{2}) - 2 (1 - x))

- 400 (y - 3 x^{2}) + 2

Solution steps

\frac{\partial}{\partial x} (- 400 x (y - x^{2}) - 2 (1 - x))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (400 x (y - x^{2})) - \frac{\partial}{\partial x} (2 (1 - x))

\frac{\partial}{\partial x} (400 x (y - x^{2})) = 400 (y - 3 x^{2})

\frac{\partial}{\partial x} (2 (1 - x)) = - 2

= - 400 (y - 3 x^{2}) - (- 2)

Simplify

= - 400 (y - 3 x^{2}) + 2

y^{^{'}} = sin (x - y)

\frac{d}{d x} (c x^{- 6})

\int (x + 1) (2 x - 1) d x

\frac{1}{x + 6}

\int \frac{1}{x ( ln ( x ) ) ^{15}} d x

What is (\partial)/(\partial x)(-400x(y-x^2)-2(1-x)) ?
The answer to (\partial)/(\partial x)(-400x(y-x^2)-2(1-x)) is -400(y-3x^2)+2