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

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

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(\partial)/(\partial x)(1/(ln(x-y)))

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 1 \sum \infty \frac{1}{n ^{0.3}}

\int - e^{- y} d y

f (x) = x^{2} \cdot sin (x)

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

y = 0.03 x^{2} + 1

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