What is (\partial)/(\partial x)(ln(kt-mx)) ?

The answer to (\partial)/(\partial x)(ln(kt-mx)) is -m/(kt-mx)

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(\partial)/(\partial x)(ln(kt-mx))

\frac{\partial}{\partial x} (ln (k t - m x))

- \frac{m}{k t - m x}

Solution steps

\frac{\partial}{\partial x} (ln (k t - m x))

Treat

k, t, m

as constants

Apply the chain rule:

\frac{1}{k t - m x} \frac{\partial}{\partial x} (k t - m x)

= \frac{1}{k t - m x} \frac{\partial}{\partial x} (k t - m x)

\frac{\partial}{\partial x} (k t - m x) = - m

= \frac{1}{k t - m x} (- m)

Simplify

\frac{1}{k t - m x} (- m) : - \frac{m}{k t - m x}

= - \frac{m}{k t - m x}

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

(x^{2})^{- 1}

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

\int \frac{1}{x ( 10 - x )} d x

\frac{\partial}{\partial x} (\frac{x + 2}{y + 1})

What is (\partial)/(\partial x)(ln(kt-mx)) ?
The answer to (\partial)/(\partial x)(ln(kt-mx)) is -m/(kt-mx)