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

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

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

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

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

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

Simplify

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

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

ma c l a u r in ln (4 + x^{2})

\int_{1}^{2} 15 4 x^{2} - 3 d x

x \to \infty lim (\frac{1}{8})

t an g e n t f (x) = \frac{3}{4 x + 1}, (- 1, - 1)

\frac{d}{d x} (2 x + \frac{200}{x})

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