What is (\partial)/(\partial x)(20x+80y-x^2-2y^2) ?

The answer to (\partial)/(\partial x)(20x+80y-x^2-2y^2) is -2x+20

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(\partial)/(\partial x)(20x+80y-x^2-2y^2)

\frac{\partial}{\partial x} (20 x + 80 y - x^{2} - 2 y^{2})

- 2 x + 20

Solution steps

\frac{\partial}{\partial x} (20 x + 80 y - x^{2} - 2 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (20 x) + \frac{\partial}{\partial x} (80 y) - \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (2 y^{2})

\frac{\partial}{\partial x} (20 x) = 20

\frac{\partial}{\partial x} (80 y) = 0

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

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

= 20 + 0 - 2 x - 0

Simplify

= - 2 x + 20

\frac{d}{d x} (sin^{7} (2 x))

\frac{d}{d x} (- 10 e^{- 2 x})

y = \frac{6 x}{7 - tan ( x )}

\int \frac{1}{1 + 2 y} d y

\int \frac{1}{x 3 - x ^{2}} d x

What is (\partial)/(\partial x)(20x+80y-x^2-2y^2) ?
The answer to (\partial)/(\partial x)(20x+80y-x^2-2y^2) is -2x+20