What is the derivative of (-4x/(x^4-2x^2+1)) ?

The derivative of (-4x/(x^4-2x^2+1)) is (4(3x^2+1))/((x^2-1)^3)

What is the first derivative of (-4x/(x^4-2x^2+1)) ?

The first derivative of (-4x/(x^4-2x^2+1)) is (4(3x^2+1))/((x^2-1)^3)

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

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

Solution steps

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

Simplify

\frac{- 4 x}{x ^{4} - 2 x ^{2} + 1} : - \frac{4 x}{x ^{4} - 2 x ^{2} + 1}

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{4} - 2 x^{2} + 1) = 4 x^{3} - 4 x

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

Simplify

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

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

x \to a lim (f (x) (x)^{g (x)})

\frac{d}{d x} (\frac{x ^{2}}{16})

\int sec^{2} (x) + tan (x) d x

\frac{d}{d x} (f (x) (3 x^{2} + 5))

\frac{\partial}{\partial x} (y^{2} - x)

What is the derivative of (-4x/(x^4-2x^2+1)) ?
The derivative of (-4x/(x^4-2x^2+1)) is (4(3x^2+1))/((x^2-1)^3)
What is the first derivative of (-4x/(x^4-2x^2+1)) ?
The first derivative of (-4x/(x^4-2x^2+1)) is (4(3x^2+1))/((x^2-1)^3)