What is the derivative of 8/(x^2+4) ?

The derivative of 8/(x^2+4) is -(16x)/((x^2+4)^2)

What is the first derivative of 8/(x^2+4) ?

The first derivative of 8/(x^2+4) is -(16x)/((x^2+4)^2)

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derivative of

\frac{8}{x ^{2} + 4}

- \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8}{x ^{2} + 4})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{1}{x ^{2} + 4})

Apply exponent rule:

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

= 8 \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

= 8 (- \frac{1}{( x ^{2} + 4 ) ^{2}} \cdot 2 x)

Simplify

8 (- \frac{1}{( x ^{2} + 4 ) ^{2}} \cdot 2 x) : - \frac{16 x}{( x ^{2} + 4 ) ^{2}}

= - \frac{16 x}{( x ^{2} + 4 ) ^{2}}

\int 9 x sec (x) tan (x) d x

y^{^{''}} + y = cos^{2} (x)

\int 3 v (v^{2} + 2)^{2} d v

\int (3 x 1 - 2 x^{2}) d x

4 x + 3, x^{3} + 3

What is the derivative of 8/(x^2+4) ?
The derivative of 8/(x^2+4) is -(16x)/((x^2+4)^2)
What is the first derivative of 8/(x^2+4) ?
The first derivative of 8/(x^2+4) is -(16x)/((x^2+4)^2)