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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{5 x ^{2} + 2 x - 6}{3 x - 1})

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

\frac{d}{d x} (ln (x) sin (x) cos (x))

f (x) = lo g_{2} (3 x)

f (x) = - 8.25 x^{2} + 66 x

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