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

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

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(\partial)/(\partial x)(1/(x+y+z))

\frac{\partial}{\partial x} (\frac{1}{x + y + z})

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

Solution steps

\frac{\partial}{\partial x} (\frac{1}{x + y + z})

Treat

y, z

as constants

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int \frac{1}{x ^{4} x ^{2} + 4} d x

(x - 3)^{- 2}

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

x \to 2 lim (\frac{x ^{3} - 2 x ^{2}}{10 x - 20})

x \to \infty lim (\frac{i}{x})

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