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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = \frac{2 x}{1 + x ^{2}}

\frac{\partial}{\partial x} (\frac{4 x ^{3}}{x ^{2} + y ^{2}})

\frac{d y}{d x} = 9 y e^{x + 9}

\frac{\partial}{\partial x} (\frac{1}{2 x})

x \to 8 lim (- (2 x - 8))

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