What is the derivative of 8/(x^2+1) ?

The derivative of 8/(x^2+1) is -(16x)/((x^2+1)^2)

What is the first derivative of 8/(x^2+1) ?

The first derivative of 8/(x^2+1) is -(16x)/((x^2+1)^2)

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

- \frac{16 x}{( x ^{2} + 1 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{16 x}{( x ^{2} + 1 ) ^{2}}

x (x^{2} + 1)^{3}

f (x) = x^{4} - 4 x^{2}, at x = 1

\int 25 - y^{2} d y

3 x - 6, - 2 x + 10, 2, 5

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

What is the derivative of 8/(x^2+1) ?
The derivative of 8/(x^2+1) is -(16x)/((x^2+1)^2)
What is the first derivative of 8/(x^2+1) ?
The first derivative of 8/(x^2+1) is -(16x)/((x^2+1)^2)