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

The derivative of (2x)/((1+x^2)^2) is (2(-3x^2+1))/((1+x^2)^3)

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

The first derivative of (2x)/((1+x^2)^2) is (2(-3x^2+1))/((1+x^2)^3)

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

2 \cdot \frac{1 \cdot ( 1 + x ^{2} ) ^{2} - 4 x ( 1 + x ^{2} ) x}{( ( 1 + x ^{2} ) ^{2} ) ^{2}} : \frac{2 ( - 3 x ^{2} + 1 )}{( 1 + x ^{2} ) ^{3}}

= \frac{2 ( - 3 x ^{2} + 1 )}{( 1 + x ^{2} ) ^{3}}

f (x) = \frac{x}{1 - ln ( x - 7 )}

cos (x), \frac{3 x}{2 π}, [0, 1]

y^{^{'}} = \frac{( ln ^{2} ( x ) )}{3 x}

y = x^{2} - 3 x

x \to - 3 - lim (h (x))

What is the derivative of (2x)/((1+x^2)^2) ?
The derivative of (2x)/((1+x^2)^2) is (2(-3x^2+1))/((1+x^2)^3)
What is the first derivative of (2x)/((1+x^2)^2) ?
The first derivative of (2x)/((1+x^2)^2) is (2(-3x^2+1))/((1+x^2)^3)