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

The derivative of (x^2+2/(x^4-3x^2+1)) is (-2x^5-8x^3+14x)/((x^4-3x^2+1)^2)

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

The first derivative of (x^2+2/(x^4-3x^2+1)) is (-2x^5-8x^3+14x)/((x^4-3x^2+1)^2)

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

\frac{- 2 x ^{5} - 8 x ^{3} + 14 x}{( x ^{4} - 3 x ^{2} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{4} - 3 x^{2} + 1) = 4 x^{3} - 6 x

= \frac{2 x ( x ^{4} - 3 x ^{2} + 1 ) - ( 4 x ^{3} - 6 x ) ( x ^{2} + 2 )}{( x ^{4} - 3 x ^{2} + 1 ) ^{2}}

Expand

2 x (x^{4} - 3 x^{2} + 1) - (4 x^{3} - 6 x) (x^{2} + 2) : - 2 x^{5} - 8 x^{3} + 14 x

= \frac{- 2 x ^{5} - 8 x ^{3} + 14 x}{( x ^{4} - 3 x ^{2} + 1 ) ^{2}}

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 4

\int x^{2} + 1 x d x

x \to 0 lim (sin (x) \cdot ln (4 x))

\frac{\partial}{\partial x} (2 x + y - 5)

\frac{d}{d x} (arccos (\frac{y}{x}))

What is the derivative of (x^2+2/(x^4-3x^2+1)) ?
The derivative of (x^2+2/(x^4-3x^2+1)) is (-2x^5-8x^3+14x)/((x^4-3x^2+1)^2)
What is the first derivative of (x^2+2/(x^4-3x^2+1)) ?
The first derivative of (x^2+2/(x^4-3x^2+1)) is (-2x^5-8x^3+14x)/((x^4-3x^2+1)^2)