What is (\partial)/(\partial y)(ln(1+xy)) ?

The answer to (\partial)/(\partial y)(ln(1+xy)) is x/(1+xy)

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(\partial)/(\partial y)(ln(1+xy))

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

\frac{x}{1 + x y}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 1 \sum \infty \frac{1}{n} - \frac{1}{2 n + 1}

1 - x^{2}, 0

\frac{d}{d x} (2 - 3 x^{2})

f (x) = \frac{x ^{2}}{1 + 4 x}

\int \frac{1}{( arcsin ( x ) ) ^{3} 1 - x ^{2}} d x

What is (\partial)/(\partial y)(ln(1+xy)) ?
The answer to (\partial)/(\partial y)(ln(1+xy)) is x/(1+xy)