What is (\partial)/(\partial x)((x^2+y^2)^{-1}) ?

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \cdot \frac{d y}{d x} = x tan (\frac{y}{x}) + y

\int x (x + 8)^{5} d x

\frac{d}{d x} (- 3 cos^{2} (x))

y = x^{3} + 1

\frac{6 x + 8 x ^{2}}{( 3 + 8 x ) ^{2}}

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