What is (\partial)/(\partial x)(ln(1-x-y)-x^{-1}-y^{-1}) ?

The answer to (\partial)/(\partial x)(ln(1-x-y)-x^{-1}-y^{-1}) is 1/(x^2)-1/(1-x-y)

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(\partial)/(\partial x)(ln(1-x-y)-x^{-1}-y^{-1})

\frac{\partial}{\partial x} (ln (1 - x - y) - x^{- 1} - y^{- 1})

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

Solution steps

\frac{\partial}{\partial x} (ln (1 - x - y) - x^{- 1} - y^{- 1})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (ln (1 - x - y)) = - \frac{1}{1 - x - y}

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

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

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

Simplify

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

f (x) = e^{- 5 x}

y = x + 10

\frac{( 5 )}{s ^{2} - 9}

t^{3} + 3 t^{2} + 3 t + 1

\frac{d}{d t} (t^{- 2} ln (t))

What is (\partial)/(\partial x)(ln(1-x-y)-x^{-1}-y^{-1}) ?
The answer to (\partial)/(\partial x)(ln(1-x-y)-x^{-1}-y^{-1}) is 1/(x^2)-1/(1-x-y)