What is the derivative of 2/(x^2+4) ?

The derivative of 2/(x^2+4) is -(4x)/((x^2+4)^2)

What is the first derivative of 2/(x^2+4) ?

The first derivative of 2/(x^2+4) is -(4x)/((x^2+4)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (t) = t^{3} e^{- 2 t} cos (6 t)

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

ln (3 x + 1), 3

\frac{d}{d x} (4 x^{2} + 5)

\frac{1}{( t ^{2} + 3 t + 1 ) ^{\frac{5}{2}}}

What is the derivative of 2/(x^2+4) ?
The derivative of 2/(x^2+4) is -(4x)/((x^2+4)^2)
What is the first derivative of 2/(x^2+4) ?
The first derivative of 2/(x^2+4) is -(4x)/((x^2+4)^2)