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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

f (x) = 4 - x - \frac{2}{3}

e^{1 + x^{2}}

2 e^{- t} cos (4 t) 1 (t)

y = ln (x + 2)

(\frac{1}{2}, - \frac{1}{2}), (5, - \frac{1}{2})

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