What is the derivative of ln(x+y-1) ?

The derivative of ln(x+y-1) is (\frac{dy)/(dx)+1}{x+y-1}

What is the first derivative of ln(x+y-1) ?

The first derivative of ln(x+y-1) is (\frac{dy)/(dx)+1}{x+y-1}

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\frac{d}{d x} (ln (x + y - 1))

\frac{\frac{d y}{d x} + 1}{x + y - 1}

Solution steps

\frac{d}{d x} (ln (x + y - 1))

Apply the chain rule:

\frac{1}{x + y - 1} \frac{d}{d x} (x + y - 1)

= \frac{1}{x + y - 1} \frac{d}{d x} (x + y - 1)

\frac{d}{d x} (x + y - 1) = \frac{d y}{d x} + 1

= \frac{1}{x + y - 1} (\frac{d y}{d x} + 1)

Simplify

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

= \frac{\frac{d y}{d x} + 1}{x + y - 1}

What is the derivative of ln(x+y-1) ?
The derivative of ln(x+y-1) is (\frac{dy)/(dx)+1}{x+y-1}
What is the first derivative of ln(x+y-1) ?
The first derivative of ln(x+y-1) is (\frac{dy)/(dx)+1}{x+y-1}