What is (\partial)/(\partial x)((7y)/(x^2+y^2+19)) ?

The answer to (\partial)/(\partial x)((7y)/(x^2+y^2+19)) is -(14yx)/((x^2+y^2+19)^2)

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(\partial)/(\partial x)((7y)/(x^2+y^2+19))

\frac{\partial}{\partial x} (\frac{7 y}{x ^{2} + y ^{2} + 19})

- \frac{14 y x}{( x ^{2} + y ^{2} + 19 ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{7 y}{x ^{2} + y ^{2} + 19})

Treat

y

as a constant

Take the constant out:

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

= 7 y \frac{\partial}{\partial x} (\frac{1}{x ^{2} + y ^{2} + 19})

Apply exponent rule:

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

= 7 y \frac{\partial}{\partial x} ((x^{2} + y^{2} + 19)^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + y ^{2} + 19 ) ^{2}} \frac{\partial}{\partial x} (x^{2} + y^{2} + 19)

= - \frac{1}{( x ^{2} + y ^{2} + 19 ) ^{2}} \frac{\partial}{\partial x} (x^{2} + y^{2} + 19)

\frac{\partial}{\partial x} (x^{2} + y^{2} + 19) = 2 x

= 7 y (- \frac{1}{( x ^{2} + y ^{2} + 19 ) ^{2}} \cdot 2 x)

Simplify

7 y (- \frac{1}{( x ^{2} + y ^{2} + 19 ) ^{2}} \cdot 2 x) : - \frac{14 y x}{( x ^{2} + y ^{2} + 19 ) ^{2}}

= - \frac{14 y x}{( x ^{2} + y ^{2} + 19 ) ^{2}}

- \frac{18}{( x - 9 ) ^{2}}

f (x) = 22 x

y = - 6 (4 x^{2} + 4)^{- 6}

\int_{0}^{\infty} arctan (x) d x

\frac{\partial}{\partial x} (x - 4)

What is (\partial)/(\partial x)((7y)/(x^2+y^2+19)) ?
The answer to (\partial)/(\partial x)((7y)/(x^2+y^2+19)) is -(14yx)/((x^2+y^2+19)^2)