What is (\partial)/(\partial x)(5/(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(5/(x^2+y^2)) is -(10x)/((x^2+y^2)^2)

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(\partial)/(\partial x)(5/(x^2+y^2))

\frac{\partial}{\partial x} (\frac{5}{x ^{2} + y ^{2}})

- \frac{10 x}{( x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{5}{x ^{2} + y ^{2}})

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

= 5 \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

= 5 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x)

Simplify

5 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x) : - \frac{10 x}{( x ^{2} + y ^{2} ) ^{2}}

= - \frac{10 x}{( x ^{2} + y ^{2} ) ^{2}}

\int_{0}^{\frac{π}{2}} e^{t} cos (3 t) d t

\frac{d}{d x} (3 arccos (\frac{x}{8}))

\frac{d y}{d x} = 3 + y - 3 x + 2

y = \frac{1}{x + 1}, y = x + 2, x = 2

\frac{d}{d x} ((x + 1)^{x})

What is (\partial)/(\partial x)(5/(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(5/(x^2+y^2)) is -(10x)/((x^2+y^2)^2)