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}}

\int 8 e^{- k x} d x

\int \frac{1}{x ^{2} + 6 x + 16} d x

x \to 0 lim (\frac{sin ( x )}{9 x})

6 x^{2} y^{^{''}} + 13 x y^{^{'}} + y = 70 x + 2

\int_{- 2}^{1} 2 x + 4 d x

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)