What is (\partial)/(\partial x)(3-x/3-y/2) ?

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

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(\partial)/(\partial x)(3-x/3-y/2)

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

- \frac{1}{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3) = 0

\frac{\partial}{\partial x} (\frac{x}{3}) = \frac{1}{3}

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

= 0 - \frac{1}{3} - 0

Simplify

= - \frac{1}{3}

f (x) = - \frac{x ^{2}}{2} - 4 x - 10

\int \frac{( 50 x + 6 )}{( 7 x + 1 ) ( x - 1 )} d x

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

g (u) = u + 1 (1 - 8 u^{2})^{7}

\frac{d}{d x} (ln (1 + \frac{ln ( 2 )}{x}))

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