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

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

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

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

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

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}}

e^{- 3 t} + 3 + 2 t + 4 t^{2} + 5 t^{3}

\frac{d}{d x} (sin (2 x) + cos (x))

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

f (x) = 5 x - x^{2}, (1, 4)

(1 + x) d y - y d x = 0

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