What is (\partial)/(\partial y)(x+y-320) ?

The answer to (\partial)/(\partial y)(x+y-320) is 1

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(\partial)/(\partial y)(x+y-320)

\frac{\partial}{\partial y} (x + y - 320)

1

Solution steps

\frac{\partial}{\partial y} (x + y - 320)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x) + \frac{\partial}{\partial y} (y) - \frac{\partial}{\partial y} (320)

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

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

\frac{\partial}{\partial y} (320) = 0

= 0 + 1 - 0

Simplify

= 1

\frac{\partial}{\partial x} (x^{4} + y^{3})

\frac{1}{1 + x}, 0

n = 1 \sum \infty \frac{- 9 ^{n}}{n ^{4}}

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

\int_{- 3}^{3} (13 - x^{2}) - (x^{2} - 5) d x

What is (\partial)/(\partial y)(x+y-320) ?
The answer to (\partial)/(\partial y)(x+y-320) is 1