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

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

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

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

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

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

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

\int \frac{( x ^{3} - 4 )}{x ^{3} + 2 x ^{2} + 2 x} d x

\frac{d}{d x} (- \frac{cos ( 2 x )}{2})

\int t^{5} ln (t) d t

e^{(e^{(x)} - x)}

f (x) = tan (4 x^{2} - 3 x - 4)

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