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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (y) = 1

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

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

y = \frac{3}{x ^{2} - 1}

x \to 0 + lim (6 (cot (x) - \frac{1}{x}))

\frac{d}{d x} (- e^{- 2 x})

f (x) = 5 x + 4, at x = 9

y^{^{''}} + 10 y^{^{'}} + 25 y = 2 r^{- 5 x}

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