What is (\partial)/(\partial x)(4/(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(4/(x^2+y^2)) is -(8x)/((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)(4/(x^2+y^2))

\frac{\partial}{\partial x} (\frac{4}{x ^{2} + y ^{2}})

- \frac{8 x}{( x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{4}{x ^{2} + y ^{2}})

Treat

y

as a constant

Take the constant out:

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

= 4 \frac{\partial}{\partial x} (\frac{1}{x ^{2} + y ^{2}})

Apply exponent rule:

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

= 4 \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

= 4 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x)

Simplify

4 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x) : - \frac{8 x}{( x ^{2} + y ^{2} ) ^{2}}

= - \frac{8 x}{( x ^{2} + y ^{2} ) ^{2}}

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

x \to 0 lim (\frac{x - 1}{x ^{2} - 1})

(x^{2} + 1) (x + 5 + \frac{1}{x})

\frac{\partial}{\partial y} (4 x e^{x^{2} + y^{2}})

y^{^{'}} = (3 x - 2)^{2} (30 x + 29)

What is (\partial)/(\partial x)(4/(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(4/(x^2+y^2)) is -(8x)/((x^2+y^2)^2)