What is (\partial)/(\partial y)(7/(x^2+y^2)) ?

The answer to (\partial)/(\partial y)(7/(x^2+y^2)) is -(14y)/((x^2+y^2)^2)

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(\partial)/(\partial y)(7/(x^2+y^2))

\frac{\partial}{\partial y} (\frac{7}{x ^{2} + y ^{2}})

- \frac{14 y}{( x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{7}{x ^{2} + y ^{2}})

Treat

x

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2}})

Apply exponent rule:

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

= 7 \frac{\partial}{\partial y} ((x^{2} + y^{2})^{- 1})

Apply the chain rule:

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

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

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

= 7 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 y)

Simplify

7 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 y) : - \frac{14 y}{( x ^{2} + y ^{2} ) ^{2}}

= - \frac{14 y}{( x ^{2} + y ^{2} ) ^{2}}

\frac{( x ^{2} + 32 x )}{8}, at x = 4

(t + 2)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

\frac{d}{d x} (f (x) \cdot \frac{1}{g ( x )})

\frac{y ^{^{'}} - e ^{- t} + 6}{y} = - 6

2 t y^{^{'}} - y = 2 t y^{3} cos (t)

What is (\partial)/(\partial y)(7/(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(7/(x^2+y^2)) is -(14y)/((x^2+y^2)^2)