What is (\partial)/(\partial x)(ln(y-x)-x/(y-x)) ?

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

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(\partial)/(\partial x)(ln(y-x)-x/(y-x))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

- \frac{1}{y - x} - \frac{y}{( y - x ) ^{2}} : \frac{- 2 y + x}{( y - x ) ^{2}}

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

\frac{d}{d x} (40320 x)

\int_{- 6}^{- 10} \frac{1}{x + 2} d x

f (x) = x^{2} - 5, (3, 4)

y = \frac{1}{x ^{7}}

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

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