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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

- sec (7 x) - ln (6 x)

x y^{^{'}} - y = x^{2017} y^{2}

e^{3 t} t^{2}

\int c^{3} t d t

(- 2, - 1), (3, 4)

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