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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \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}}

= 2 \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

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

Simplify

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

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

\frac{d}{d x} ((x - 4)^{3} e^{- 4 x})

\frac{d y}{d x} = e^{3 x + 5 y}

\int 4 tan^{3} (x) sec^{2} (x) d x

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

\int \frac{1}{x ^{2}} - \frac{4}{x ^{3}} d x

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