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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\frac{d}{d x} (\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}}

\int \frac{x ^{4}}{x ^{4} - 16} d x

y^{^{''}} + 9 y = 2 cos (3 x) + 3 sin (3 x)

f (x) = 4 sin (x), at x = \frac{π}{2}

y y^{^{'}} = y^{2} + 1

f (x) = csc^{3} (5 x^{2} + 1)

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)