What is (\partial)/(\partial y)(ln(x^2+y^2+2)) ?

The answer to (\partial)/(\partial y)(ln(x^2+y^2+2)) is (2y)/(x^2+y^2+2)

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(\partial)/(\partial y)(ln(x^2+y^2+2))

\frac{\partial}{\partial y} (ln (x^{2} + y^{2} + 2))

\frac{2 y}{x ^{2} + y ^{2} + 2}

Solution steps

\frac{\partial}{\partial y} (ln (x^{2} + y^{2} + 2))

Treat

x

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (x^{2} + y^{2} + 2) = 2 y

= \frac{1}{x ^{2} + y ^{2} + 2} \cdot 2 y

Simplify

\frac{1}{x ^{2} + y ^{2} + 2} \cdot 2 y : \frac{2 y}{x ^{2} + y ^{2} + 2}

= \frac{2 y}{x ^{2} + y ^{2} + 2}

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

\int (1 + x)^{3} d x

f (x) = \frac{x ^{2} - 3 x + 4}{x + 2}

1 + x - x^{2} - x^{3}, 1 - x

\frac{d}{d x} (\frac{6}{2 x + 1})

What is (\partial)/(\partial y)(ln(x^2+y^2+2)) ?
The answer to (\partial)/(\partial y)(ln(x^2+y^2+2)) is (2y)/(x^2+y^2+2)