What is (\partial)/(\partial y)(ln(x^2+y^2+1)) ?

The answer to (\partial)/(\partial y)(ln(x^2+y^2+1)) is (2y)/(x^2+y^2+1)

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(\partial)/(\partial y)(ln(x^2+y^2+1))

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (3 - \frac{e ^{y}}{e ^{x + 1}})

f (x) = 2 x^{2} + 3 x - 2, at x = 0

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

x \to 1 + lim (x^{3})

What is (\partial)/(\partial y)(ln(x^2+y^2+1)) ?
The answer to (\partial)/(\partial y)(ln(x^2+y^2+1)) is (2y)/(x^2+y^2+1)