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

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

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(\partial)/(\partial x)(y/(x^2+y^2+z^2))

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

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

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = 7 x + 1, (5, 6)

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

\frac{\partial}{\partial x} (A + B \cdot cos (x) sin (y))

\frac{d x}{d t} = x^{\frac{2}{3}}

\frac{1}{4 s ^{2}}

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