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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

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

Simplify

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

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

\frac{a - x}{a + x}

x \to 2 lim (x^{2} + 2 x)

x \to 0 + lim \frac{1}{4 x ^{\frac{6}{7}}} - \frac{7}{( x - 6 ^{\frac{6}{7}} )}

\frac{12}{x ^{5}}

\frac{9 - sec ( x )}{tan ( x )}

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