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

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

Popular >

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int x^{2} + 5 x - 6 d x

\frac{d}{d x} (2 \cdot 1 0^{- 3} x^{3} + 10 x)

\frac{\partial}{\partial s} (\frac{2 r + s}{t})

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

3 x^{2} + 7 x - 4

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