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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 30 (x + 7) (x + 5)^{4} d x

\frac{d}{d x} (ln (\frac{x}{4}))

n = 1 \sum \infty \frac{x ^{5 n}}{n !}

f (x) = x e^{9 x}

x \to 8 lim (\frac{8 x}{x - 8})

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