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

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

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

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

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derivative of

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

t \cdot y^{^{'}} + 2 \cdot y = t^{2} - t + 1

f (x) = e^{- 3 x} sin (4 x^{2})

\int e^{9 - 4 x} d x

\int \frac{cos ( x )}{3 + 2 sin ( x )} d x

y = 6 x (x^{2} + 2) 2

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