What is (\partial)/(\partial x)(1/(x^2+4)) ?

The answer to (\partial)/(\partial x)(1/(x^2+4)) is -(2x)/((x^2+4)^2)

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(\partial)/(\partial x)(1/(x^2+4))

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4})

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

Solution steps

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4})

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{( x ^{2} + 4 ) ^{2}} \frac{\partial}{\partial x} (x^{2} + 4)

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

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

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

Simplify

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

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

\frac{d y}{d x} = \frac{x ^{3}}{y + 1}

x y^{^{'}} - y = x^{2} + y^{2}

x \to 2 lim (5 x^{3} - 3 x^{2} + x - 6)

e^{x}, x e^{x}, x = 0

\frac{d}{d x} (3 e^{- \frac{3}{x}})

What is (\partial)/(\partial x)(1/(x^2+4)) ?
The answer to (\partial)/(\partial x)(1/(x^2+4)) is -(2x)/((x^2+4)^2)