What is (\partial)/(\partial x)(8/(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(8/(x^2+y^2)) is -(16x)/((x^2+y^2)^2)

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(\partial)/(\partial x)(8/(x^2+y^2))

\frac{\partial}{\partial x} (\frac{8}{x ^{2} + y ^{2}})

- \frac{16 x}{( x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{8}{x ^{2} + y ^{2}})

Treat

y

as a constant

Take the constant out:

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

= 8 \frac{\partial}{\partial x} (\frac{1}{x ^{2} + y ^{2}})

Apply exponent rule:

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

= 8 \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

= 8 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x)

Simplify

8 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x) : - \frac{16 x}{( x ^{2} + y ^{2} ) ^{2}}

= - \frac{16 x}{( x ^{2} + y ^{2} ) ^{2}}

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

x - 1

\frac{\partial}{\partial y} (x^{2} \cdot sin (2 y))

f (x) = x - 3 x

f (x) = x^{2} + x - 9

What is (\partial)/(\partial x)(8/(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(8/(x^2+y^2)) is -(16x)/((x^2+y^2)^2)