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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

= - \frac{1 \cdot ( y ^{2} + x ^{2} ) - 2 yy}{( y ^{2} + x ^{2} ) ^{2}}

1 \cdot (y^{2} + x^{2}) - 2 yy = - y^{2} + x^{2}

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 t e^{3 t}

\int_{0}^{6} 36 - x^{2} d x

x \to 0 + lim (ln (ln (x)))

\frac{d y}{d x} = \frac{1}{sec ^{2} ( y )}

\frac{d}{d x} (\frac{x ^{4} + 2}{ln ( x )})

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