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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 lim (\frac{x ^{2}}{x ^{4}})

\frac{d}{d x} (x^{3} - 3 x^{2} - 8 x + 2)

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

\frac{\partial}{\partial x} (x^{2} cos (x y + 2 x z - 3 yz))

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

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