What is (\partial)/(\partial x)(xy+1/x+1/y) ?

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

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(\partial)/(\partial x)(xy+1/x+1/y)

\frac{\partial}{\partial x} (x y + \frac{1}{x} + \frac{1}{y})

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

Solution steps

\frac{\partial}{\partial x} (x y + \frac{1}{x} + \frac{1}{y})

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{1}{x}) + \frac{\partial}{\partial x} (\frac{1}{y})

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

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

\frac{\partial}{\partial x} (\frac{1}{y}) = 0

= y - \frac{1}{x ^{2}} + 0

Simplify

= y - \frac{1}{x ^{2}}

f (x) = 9 - 6 x, at x = - 1

- \frac{3}{s + 3} + \frac{4}{s - 4}

\int \frac{8}{x ^{- 1} + 2} d x

y = 4 x^{2} - x

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

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