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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{d}{d x} (ln (4) lo g_{4} (x) + π^{3})

x \to 2 lim (3.9 + 9.6)

n = 0 \sum \infty \frac{10}{2 ^{n}}

\frac{d}{d x} (\frac{x}{6 x + 5})

\frac{d}{d t} (sin (t) \cdot cos (t))

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