What is (\partial)/(\partial x)(x/(x^2+4)) ?

The answer to (\partial)/(\partial x)(x/(x^2+4)) is (-x^2+4)/((x^2+4)^2)

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(\partial)/(\partial x)(x/(x^2+4))

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 4})

\frac{- x ^{2} + 4}{( x ^{2} + 4 ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 4})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{\partial}{\partial x} ( x ) ( x ^{2} + 4 ) - \frac{\partial}{\partial x} ( x ^{2} + 4 ) x}{( x ^{2} + 4 ) ^{2}}

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (x^{2} + 4) = 2 x

= \frac{1 \cdot ( x ^{2} + 4 ) - 2 xx}{( x ^{2} + 4 ) ^{2}}

1 \cdot (x^{2} + 4) - 2 xx = - x^{2} + 4

= \frac{- x ^{2} + 4}{( x ^{2} + 4 ) ^{2}}

x \frac{d v}{d x} = \frac{1 - 4 v ^{2}}{3 v}

(x^{2} + y^{2} - 7) d x = (y + x y) d y, y (0) = 1

\frac{d}{d x} (lo g_{3} (\frac{3 x ^{2}}{x}))

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

\int cos (4 u) d u

What is (\partial)/(\partial x)(x/(x^2+4)) ?
The answer to (\partial)/(\partial x)(x/(x^2+4)) is (-x^2+4)/((x^2+4)^2)