What is the derivative of 1/(x^2+4) ?

The derivative of 1/(x^2+4) is -(2x)/((x^2+4)^2)

What is the first derivative of 1/(x^2+4) ?

The first derivative of 1/(x^2+4) is -(2x)/((x^2+4)^2)

derivative of

\frac{1}{x ^{2} + 4}

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

Solution steps

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

Apply exponent rule:

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

= \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

= - \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}{d x} (\frac{tan ( x )}{x ^{2}})

t an g e n t f (x) = \frac{4 x}{x + 3}, (1, 1)

\frac{d}{d θ} (cos (θ) - θ sin (θ))

a re a 9 x^{2}, 8 x^{4} - 8 x^{2}

\frac{d y}{d x} + 2 x^{2} = 0

What is the derivative of 1/(x^2+4) ?
The derivative of 1/(x^2+4) is -(2x)/((x^2+4)^2)
What is the first derivative of 1/(x^2+4) ?
The first derivative of 1/(x^2+4) is -(2x)/((x^2+4)^2)