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)

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(\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}}

\frac{d}{d x} (x^{3} + e^{2 x})

x \to 2 lim (x^{2} + 3 x - 1)

\frac{\partial}{\partial x} (\frac{4 x - 6 y}{y ^{2} + 2 x ^{2}})

\frac{d}{d x} (sin (ln (cos (x))))

\frac{d}{d x} (\frac{5 x ^{2} - 9 x + 13}{2 x + 4})

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)