What is (\partial)/(\partial x)(xy-2/x-4/y+8) ?

The answer to (\partial)/(\partial x)(xy-2/x-4/y+8) is 2/(x^2)+y

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(\partial)/(\partial x)(xy-2/x-4/y+8)

\frac{\partial}{\partial x} (x y - \frac{2}{x} - \frac{4}{y} + 8)

\frac{2}{x ^{2}} + y

Solution steps

\frac{\partial}{\partial x} (x y - \frac{2}{x} - \frac{4}{y} + 8)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial x} (\frac{2}{x}) - \frac{\partial}{\partial x} (\frac{4}{y}) + \frac{\partial}{\partial x} (8)

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

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

\frac{\partial}{\partial x} (\frac{4}{y}) = 0

\frac{\partial}{\partial x} (8) = 0

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

Simplify

= \frac{2}{x ^{2}} + y

\frac{\partial}{\partial v} (u + v)

\frac{x}{x ^{2} 1}

\frac{d}{d x} (- \frac{2}{( 1 + x ) ^{\frac{3}{2}}})

f (x) = \frac{7}{x + 1}, at x = 7

\int cos (x) cos (y) d x

What is (\partial)/(\partial x)(xy-2/x-4/y+8) ?
The answer to (\partial)/(\partial x)(xy-2/x-4/y+8) is 2/(x^2)+y